Lattice Equable Quadrilaterals I -- Parallelograms
aa r X i v : . [ m a t h . N T ] J un LATTICE EQUABLE PARALLELOGRAMS
CHRISTIAN AEBI AND GRANT CAIRNS ∗ Abstract.
This paper studies equable parallelograms whose vertices lie onthe integer lattice. Using Rosenberger’s Theorem on generalised Markovequations, we show that the g.c.d. of the side lengths of such parallelogramscan only be 3, 4 or 5, and in each of these cases the set of parallelogramsnaturally forms an infinite tree all of whose vertices have degree 4, bar theroot. The paper then focuses on what we call Pythagorean equable paral-lelograms. These are lattice equable parallelograms whose complement ina circumscribing rectangle consists of two Pythagorean triangles. We provethat for these parallelograms the shortest side can only be 3, 4, 5, 6 or 10,and there are five infinite families of such parallelograms, given by solutionsto corresponding Pell-like equations. Introduction
A polygon with integer sides is said to be equable if its perimeter equals its area.Equable polygons have fascinated recreational mathematical amateurs at least asfar back as Ozanam [7]:
Probl`eme XXIII.
D´ecrire un triangle rectangle dont l’aire en nombres soit ´egaleau contour. Probl`eme XXVII.
D´ecrire un parall´elogramme rectangle dont l’aire en nombressoit ´egale au contour. These exercises can easily be done in high school today, since they rely merelyon the ability of completing a rectangle , instead of the usual square .(a) For the Pythagorean triangle with side lengths b < c < √ b + c we have12 bc = b + c + p b + c = ⇒ bc − b − c ) − b + c )= bc ( bc − b − c + 8)= ⇒ ( b − c −
4) = 8 , and hence ( b, c ) = (5 ,
12) and (6 ,
8) are the only two possibilities.(b) For a rectangle with integer side lengths b, c with b ≤ c we have bc = 2( b + c ) ⇐⇒ bc − b − c = ( b − c − − ⇐⇒ ( b − c −
2) = 4 , giving ( b, c ) = (4 ,
4) and (3 , ∗ corresponding author Determine which right triangles have the same area as perimeter. Determine which rectangles have the same area as perimeter. ∗ The striking feature of equable integer sided triangles is that apart from thetwo Pythagorean triangles given above, there are only 3 other possibilities; seethe Appendix. In this paper we study equable parallelograms. Unlike equabletriangles, at first sight equable parallelograms are sadly disappointing, as they arejust far too common. Indeed, it is easier to itemise the non-equable cases.
Proposition 1. If b, c are positive integers with b ≤ c , then there is an equableparallelogram with sides b, c unless one of the following holds: (a) b = 1 or , and c arbitrary with b ≤ c . (b) b = 3 and c = 3 , or .Proof. Repeating the argument we used above for Ozanam’s Probl`eme XXVII, thearea is less than the perimeter when bc < b + c ) ⇐⇒ > ( b − c − , and this condition is satisfied for precisely for the values of b, c itemised in thepossible conditions of the proposition. So in these cases, there is no equable par-allelogram with side lengths b, c . Conversely, if bc ≥ b + c ), start with therectangle with side lengths b, c , and hence area bc , and gradually push the paral-lelogram over, while maintaining its side lengths, as in Figure 1. Ultimately, whenthe parallelogram becomes flat, the area is zero. So by continuity, somewhere inthe process the area equals 2( b + c ) and the parallelogram is equable. (cid:3) b b bc c c Figure 1.
Collapsing a parallelogramGiven the above, to make life interesting we restrict ourselves to consideringa subclass of equable parallelograms that has a wealth of interesting members,without becoming mundane.
Definition 1. A lattice equable parallelogram (or LEP , for short) is a parallelo-gram whose perimeter equals its area and whose vertices lie on the integer lattice Z .Notice that in the above definition, we don’t require that the side lengths beintegers. Indeed, as we show in Lemma 1 below, this fact can be deduced fromthe other hypotheses. Throughout this paper we will denote the side lengths bythe symbols b and c ( b for base and c for cˆot´e ). Moreover, for brevity, we will attimes drop the word length and simply refer to b, c as the sides . Note that a LEPis completely determined, up to a Euclidean motion, by its sides b, c . Indeed, if θ denotes one of the angles between the sides, then the area is bc sin θ and so by ATTICE EQUABLE PARALLELOGRAMS 3 equability, sin θ = 2( b + c ) /bc is determined by b and c . So our main aim is thispaper in to study the values of b, c for which a LEP exists with sides b, c .In Section 2 we use a result of Paul Yiu on Heronian triangles to prove thefollowing criteria. Theorem 1.
Given positive integers b, c , a lattice equable parallelogram with sides b, c exists if and only if b c − b + c ) is a square. In Section 3 we use Rosenberger’s Theorem on generalised Markov equationsto prove the following result.
Theorem 2.
If a lattice equable parallelogram has sides b, c , then gcd( b, c ) is 3, 4or 5. In Section 4 we describe how all LEPs can be derived from three fundamentalexamples by successive applications of four functions. This gives the forest of LEPs,consisting of three trees corresponding to the three possible values of gcd( b, c ): seeFigures 6 and 7. We investigate various aspects and applications of these results inSections 5 and 6. Then in Section 7 we analyse a large natural family of LEPs thatwe call
Pythagorean equable parallelograms . These are LEPs whose complement ina circumscribing rectangle consists of two Pythagorean triangles; see Definition 3.We use Theorem 2 to prove the following:
Theorem 3.
If a Pythagorean equable parallelogram has sides b, c with b ≤ c then b = 3 , , , or . There are 5 infinite families of such parallelograms, which aregiven by the solutions of corresponding Pell or Pell-like equations: (F1) b = 3 , c = x + y )2 , where y − x = 4 . (F2) b = 4 , c = 4( x + y ) , where y − x = 1 . (F3) b = 5 , c = x + y , where y − x = 20 . (F4) b = 6 , c = 3( x + y ) , where y − x = 1 . (F5) b = 10 , c = x + y , where y − x = 5 . At the end of Section 7 we locate the Pythagorean equable parallelograms oncertain branches of the trees of LEPs given in Figures 6 and 7.The paper concludes with an appendix that revisits the classic theorem onequable triangles that dates to 1904.2.
LEPs: general properties and special cases
Let us begin with a trivial remark that we will use on several occasions.
Remark . If the square root of an integer is rational, then it is an integer. Con-sequently, if the distance d between two integer lattice points is rational, then d isan integer. Lemma 1.
LEPs have integer side lengths.Proof.
Consider a LEP P with vertices O (0 , , A ( x, y ) , B ( z, w ) , C ( u, v ), in anti-clockwise order, where z = x + u, w = y + v . Let b denote the length of OA and c the length of OC . The area of P is xv − yu , which is an integer. By the equabilityhypothesis, 2( b + c ) is an integer. So, as b , c are integers, b − c = ( b − c ) / ( b + c ) CHRISTIAN AEBI AND GRANT CAIRNS ∗ is rational. Thus b = b + c + b − c is rational, and hence b is an integer, by Remark 1.So c is also rational, and hence an integer. (cid:3) Proposition 2.
No LEP can be partitioned into two congruent right triangles.Proof.
Suppose a LEP P can be partitioned into two congruent right triangles.Then P can be cut in half and recombined to form an equable isosceles trianglewith integer sides, as in Figure 2. The proposition then follows from the factthat the complete list of 5 equable triangles includes no isosceles triangles; see theAppendix below. b hc b hc Figure 2.
A particular non-equable parallelogramA simple and more direct proof of the proposition is as follows. The equabilityhypothesis gives hb = 2 b + 2 p b + h = ⇒ b ( h − = 4( b + h ) ⇐⇒ b h − b − h ⇐⇒ ( b − h −
4) = 16 . (1)Consequently h is rational and so by Remark 1, h is an integer. Then by (1), b − b . (cid:3) Lemma 2.
Suppose a LEP P has sides b, c . Then the lengths of the diagonals of P are given by the following formula: d = ( b + c ) ± p b c − b + c ) . In particular, b c − b + c ) is a square.Proof. Consider a diagonal of length d . By Heron’s formula, the triangle withsides b, c, d has area14 p ( b + c + d )( − b + c + d )( b − c + d )( b + c − d ) . Hence the equability hypothesis is ( b + c + d )( − b + c + d )( b − c + d )( b + c − d ) =16( b + c ) . Rearranging, this gives d − b + c ) d + ( b − c ) + 16( b + c ) = 0,so d = ( b + c ) ± p ( b + c ) − ( b − c ) − b + c ) . Simplifying, we have d = ( b + c ) ± p b c − b + c ) , as required. In particular, as b, c, d are integers, 4 b c − b + c ) is a square. (cid:3) Remark . Suppose a LEP P has sides b, c and diagonals d , d . Then the abovelemma gives d d = ( b + c ) p
16 + ( b − c ) . ATTICE EQUABLE PARALLELOGRAMS 5
Proof of Theorem 1.
The necessity of the condition was shown in Lemma 2. There-fore, assume that b, c are positive integers such that 4 b c − b + c ) is a square.Consider a triangle T with sides b, c and d := q b + c + 2 p b c − b + c ) . No-tice that such a triangle exists because b, c < d < b + c , where the latter inequalityholds as b + c + 2 p b c − b + c ) < ( b + c ) since p b c − b + c ) < bc . Let θ denote the angle between sides b, c and note that θ is obtuse since d ≥ b + c .So d = b + c − bc cos θ = b + c + 2 p b c − b c sin θ. So, from the definition of d , we have bc sin θ = 2( b + c ). Hence, since bc sin θ istwice the area of T , the area of T is b + c . Now consider the parallelogram P madefrom two copies of T . From what we have just seen, P is equable. It remains toshow that P can be realised as a lattice parallelogram, or equivalently, that T canbe realised as a lattice triangle. But since b, c are integers, and d is an integer,and the area of T is an integer, T is geodetic , in the the terminology of Paul Yiu.Thus T can be realised as a lattice triangle; see the last paragraph of [12]. (cid:3) Remark . Observe that LEPs are not determined up to congruence by their area.Indeed, by Theorem 1, there are LEPs with ( b, c ) = (3 , , (5 , , (25 ,
65) thateach have area 180. A larger example is given by ( b, c ) = (85 , , Corollary 1. (a)
The only rhombi that are LEPs are the × square and the rhombus withside length 5 and area 20. (b) If a LEP has sides b, c with c = 2 b , then b = 3 or .Proof. From Theorem 1, a rhombus with side length b is a LEP if and only if b − b is a square, that is, if b −
16 is a square. But obviously this only occurswhen b is 4 or 5. The first case is the 4 × c = 2 b , then 4 b − b is a square and so b − b = 3 or 5. (cid:3) − − Figure 3.
Equable RhombusWe conclude this section with an elementary result whose proof will hopefullygive the reader a better feel for the nature of LEPs.
CHRISTIAN AEBI AND GRANT CAIRNS ∗ Theorem 4.
There are only three LEPs lying in the first quadrant with one vertexat the origin and a diagonal lying on the line x = y . They are the × square, theLEP with vertices (0 , , (3 , , (12 , , (9 , , and its reflection in the line y = x .Proof. Consider a parallelogram P with vertices O (0 , , A ( x, y ) , B ( z, z ) , A ′ ( z − x, z − y ), in clock-wise order, where x, y, z are non-negative integers and y < x ≤ z ,as in diagram on the left of Figure 4. Let us first calculate the area α of the triangle OAB . The diagonal OB has length √ z and the distance of A to the diagonal is( x − y ) / √
2. So α = z ( x − y ). The sum σ of the lengths OA and AB is(2) σ = p x + y + p ( z − x ) + ( z − y ) . Moreover, σ is no greater than the sum of the lengths of the segments 0( z,
0) and( z, B . That is, σ ≤ z . Assume that P is a LEP. The equable hypothesis, α = σ ,gives(3) 12 z ( x − y ) ≤ z, and hence y < x ≤ y + 4. So we have 4 cases to consider: ( z, O (0 , A ( x, y ) B ( z, z ) A ′ ( z − x, z − y ) Figure 4.
Parallelograms with diagonal on line y = x Case x = y + 4. Here (15) is an equality, from which we conclude that the givenparallelogram is a square, with side length z . So, by equability, z = 4 z and hence z = 4. So A = (4 , , B = (4 , x = y + 3. Substituting this in (2), equability gives32 z = p x + ( x − + p ( z − x ) + ( z − x + 3) . Writing this as 3 z − p x + ( x − = 2 p ( z − x ) + ( z − x + 3) and squaringboth sides and simplifying gives 9 z − z p x + ( x − = 4(2 z − z (2 x − z gives z = 4(6 − x +3 p x + ( x − ). In particular, ATTICE EQUABLE PARALLELOGRAMS 7 x + ( x − is a square. Moreover, x ≤ z gives 4(6 − x + 3 p x + ( x − ) ≥ x and so 144( x + ( x − ) ≥ (17 x − and simplifying,0 ≥ x + 48 x −
720 = ( x + 60)( x − . Thus x ≤
12. Calculations show that the only such values for which x + ( x − is a square are x = 3 and x = 12. These values correspond to the parallelogram(0 , , (3 , , (12 , , (9 , y = x ; see the diagramon the right of Figure 4.Cases x = y + 2 and x = y + 1. The area is respectively z and z/ σ of the sides is greater than the length of the diagonal, so σ > √ z . Thus equability is impossible in both cases. (cid:3) Remark . The above result does not hold without the assumption that the LEPslies in the first quadrant. For example, consider the following 6 LEPs with vertices O (0 , , A ( x, y ) , B ( z, z ) , A ′ ( z − x, z − y ), in clock-wise order: A = (0 , − , B = (60 , ,A = (0 , − , B = (12 , ,A = ( − , − , B = (68 , ,A = ( − , − , B = (348 , ,A = ( − , − , B = (2028 , ,A = ( − , − , B = (396 , . Notice incidentally that in all these cases x − y ≤
4, as in the proof of the abovetheorem, even though in these cases the diagonal on y = x is the short diagonalof the LEP. This is a general fact; see Theorem 5 below.3. The LEP restriction: gcd( b, c ) = 3 , , ax + by + cz = dxyz, where a, b, c are pairwise relatively prime positive integers with a ≤ b ≤ c suchthat a, b, c all divide d . We are only interested in positive integer solutions, thatis, x, y, z ∈ N , so we use the word solution to mean positive integer solution.Rosenberger’s remarkable result is that only 6 such equations have a solution andwhen such a solution exists, there are infinitely many solutions. We use the R1–R5notation of [1]. Rosenberger’s Theorem ([9]) . Equation (4) only has a solution in the following6 cases: M: x + y + z = 3 xyz (Markov’s equation), R1: x + y + 2 z = 4 xyz , R2: x + 2 y + 3 z = 6 xyz , R3: x + y + 5 z = 5 xyz , R4: x + y + z = xyz , R5: x + y + 2 z = 2 xyz , CHRISTIAN AEBI AND GRANT CAIRNS ∗ Remark . As mentioned in [9], it is easy to see by considering R4 modulo 3, thatif ( x, y, z ) is a solution to R4, then x, y, z are each divisible by 3, and ( x, y, z )is a solution to Markov’s equation if and only if (3 x, y, z ) is a solution to R4.Similarly, by considering R5 modulo 4, if ( x, y, z ) is a solution to R5, then x, y, z are each even, and ( x, y, z ) is a solution to R1 if and only if (2 x, y, z ) is a solutionto R5.Returning to LEPs, suppose that b, c are positive integers such that b c − b + c ) is a square. Using the standard characterisation of Pythagorean triples we haverelatively prime positive integers m, n and a positive integer k such that either(5) bc = k ( m + n ) , b + c ) = 2 kmn, p b c − b + c ) = k ( n − m ) , or(6) bc = k ( m + n ) , p b c − b + c ) = 2 kmn, b + c ) = k ( n − m ) . We will show that the second possibility can be reduced to the first. Indeed, sup-pose that (6) holds. Assume for the moment that k is odd. Since p b c − b + c ) =2 kmn , it follows that bc is even. Consider the equation bc = k ( m + n ). As bc iseven and k is odd and m, n are relatively prime, m, n must both be odd. Thus,from bc = k ( m + n ), we have bc ≡ b, c is even. Thus2( b + c ) ≡ m, n are all odd, n − m ≡ b + c ) = k ( n − m ). Thus k is even, k = 2 κ say. Let y = m + n, x = n − m , so n = y + x , m = y − x . Then (6) can be written bc = k ( m + n ) = κ ( x + y ) , b + c ) = k ( n − m ) = 2 κxy, kmn = κ ( y − x ) , which has the same form as (5). Since m, n are relatively prime, gcd( x, y ) = 1 or2. In the later case we can replace x and y by x = x/ y = y/ κ by κ = 4 κ . We then have the same form as (5) again but now x, y are relatively prime. We may therefore assume (5) for what follows.Note that b, c are solutions to the equation(7) x − kmnx + k ( m + n ) = 0 . In particular, we have(8) km + kn + b = kbmn. Let k = f s , where f is square-free. Lemma 3. k is either 5, 8 or 9.Proof. From (8), f divides b and hence f divides b . Let b = f β . Dividing (8) by f gives(9) s m + s n + f β = f s mnβ. From (9), s divides f β and hence s divides β , and thus s divides β . Let β = sq .Dividing (9) by s gives(10) m + n + f q = fs mnq. Thus by Rosenberger’s Theorem, using x = m, y = n, z = q , there are five possi-bilities: ATTICE EQUABLE PARALLELOGRAMS 9 (a) f = 1 , s = 3 (corresponding to M),(b) f = 2 , s = 2 (corresponding to R1),(c) f = 5 , s = 1 (corresponding to R3),(d) f = 1 , s = 1 (corresponding to R4),(e) f = 2 , s = 1 (corresponding to R5).Cases (a) to (e) give k = 9 , , , , k = 1, then (10) is m + n + q = mnq . But then as noted in Remark 5, thevalues m, n, q are all divisible by 3, which contradicts our assumption that m, n are relatively prime. Similarly, if k = 2, then (10) is m + n + 2 q = 2 mnq . Butthen as noted in Remark 5, the values m, n, q are all even, again contradicting ourassumption that m, n are relatively prime. (cid:3) Proof of Theorem 2.
Suppose as before that bc = k ( m + n ) , b + c ) = 2 kmn, p b c − b + c ) = k ( n − m ) , so from the previous lemma, k is either 5, 8 or 9. Lemma 4.
The numbers k and gcd( b, c ) have the same prime divisors.Proof. From (7), b − kmnb + k ( m + n ) = 0 and c − kmnc + k ( m + n ) = 0.So if p is any prime divisor of k , then p divides b and c , so p divides b and c .Hence p divides gcd( b, c ).Conversely, let p be any prime divisor of gcd( b, c ), and suppose that p doesn’tdivide k . Then p divides m + n and p divides mn so p divides ( m + n ) and( m − n ) , so p divides m + n and m − n . Hence p divide 2 m and 2 n . If p were oddwe would have that p divides m and n , which is impossible as m, n are relativelyprime. So p = 2; that is, b, c are both even. Since p doesn’t divide k , we have that k is odd and m + n is even. So, as m, n are relatively prime, m, n are both odd.But then b + c = kmn implies that k is even, a contradiction. So p divides k . (cid:3) Consider the case where k = 5. From Lemma 4, gcd( b, c ) = 5 i for some i ≥ i >
1, so b, c are both divisible by 25. From bc = k ( m + n ) we havethat m + n is divisible by 5 . From b + c = kmn we have that mn is divisible by5, so exactly one of m or n is divisible by 5 since m, n are relatively prime. But thisis a contradiction since m + n would not be divisible by 5. Hence gcd( b, c ) = 5.Now suppose k = 9. From Lemma 4, gcd( b, c ) = 3 i for some i ≥
1. Supposethat i >
1, so b, c are both divisible by 9. From bc = k ( m + n ) we have that m + n is divisible by 9. In Z the squares are 0,1,4,7. Because m, n are relativelyprime, they are not both congruent to 0 modulo 3. So modulo 9, m + n ∈ { , , , , , , , , } ≡ { , , , , , } , contradicting the fact that m + n is divisible by 9. Hence gcd( b, c ) = 3.Finally, suppose k = 8. From Lemma 4, gcd( b, c ) = 2 i for some i ≥
1. As2 divides ( b + c ) = k m n and 2 divides 4 bc = 4 k ( m + n ), thus 2 divides( b + c ) − bc = ( b − c ) . Consequently, as ( b − c ) is a square, 2 divides ( b − c ) .Hence, 2 divides b − c . Thus, as 2 divides b − c and b + c = kmn , so 2 dividesboth 2 b and 2 c . Hence 2 divides both b and c , and hence gcd( b, c ) ≥ . Thus ∗ i ≥
2. Now suppose i >
2. Thus 2 divides bc = 8( m + n ) and hence 8 divides m + n . In Z the squares are 0,1,4. Because m, n are relatively prime, they arenot both even. So modulo 8, m + n ∈ { , , } ≡ { , , , } , contradicting the fact that m + n is divisible by 8. Hence gcd( b, c ) = 2 .This completes the proof of Theorem 2. (cid:3) Remark . In the case gcd( b, c ) = 3, we have k = 3 and b + c = kmn , so b + c isdivisible by 3 . Then neither b nor c is divisible by 3 since otherwise both b and c would be divisible by 3 , contradicting gcd( b, c ) = 3. Similarly, if gcd( b, c ) = 4,then b + c is divisible by 2 but neither b nor c is divisible by 2 . But a similarresult does not hold in the gcd( b, c ) = 5 case. For example, for the LEP with b = 85 , c = 1525 one has b = 5 · , c = 5 · b, c , we have gcd( b, c ) = 3 , b, c are determined by the corresponding Markov-Rosenbergerequations M,R1 and R3, as follows:(a) gcd( b, c ) = 3: m + n + q = 3 mnq , where q = b/ b, c ) = 4: m + n + 2 q = 4 mnq , where q = b/ b, c ) = 5: m + n + 5 q = 5 mnq , where q = b/ bc = k ( m + n ) , b + c = kmn and k = 9 , , m, n, q ) is asolution to one of the above equations, then by Rosenberger’s Theorem, the triple m, n, q is pairwise relatively prime. In particular, m, n are relatively prime.4. The forest of LEPs
For a given k (= 5 , S k of solutions of the correspondingMarkov-Rosenberger equation. Let us briefly recall Rosenberger’s theory. For thereader’s convenience, we recall (10): m + n + f q = fs mnq. Following the presentation given in [1], from a solution x = ( m, n, q ) to (10), onecan generate three new solutions by applying the involutions: φ ( x ) = ( f snq − m, n, q ) φ ( x ) = ( m, f smq − n, q ) φ ( x ) = ( m, n, smn − q ) . The group of transformations of S k generated by the maps φ i is the free productof three copies of Z , and this group acts transitively on S k . Moreover, the maps φ i give the set S k of solutions the structure of an infinite binary tree: each solutionis a vertex and two solutions are connected by an edge if one of the maps φ i sendsone solution to the other. The fundamental solutions have the smallest values of m + n + q ; in M,R1,R3 they are ( m, n, q ) = (1 , , , (1 , , , (1 , ,
1) respectively.Having recalled Rosenberger’s theory, we now describe how the solution trees S k of the Markov-Rosenberger equations determine the induced structure on theset T gcd of LEPs, where gcd = 3 , , k = 9 , , T gcd ATTICE EQUABLE PARALLELOGRAMS 11 as the set of ordered pairs ( b, c ) of possible side lengths with gcd( b, c ) = gcd and b ≤ c . A solution ( m, n, q ) to the corresponding Markov-Rosenberger equationcorresponds to a LEP if n ≥ m . Let τ denote the involution of S k given by τ : ( m, n, q ) ( n, m, q ). The map τ commutes with φ , and conjugates φ to φ . We form the quotient tree ¯ S k = S k /τ . The elements of ¯ S k can be regardedas triples ( m, n, q ) with m ≤ n . The map φ induces an involution ¯ φ on ¯ S k .Note that by definition φ leaves m and n unchanged and sends q to smn − q , orequivalently, b to kmn − b = b + c − b = c . Thus ( m, n, q ) and φ ( m, n, q ) correspondto the same LEP. Consequently we form T gcd by contracting each of the edges of¯ S k that are given by the map ¯ φ ; see Figure 5. Note that contraction of edges of atree produces another tree. So T gcd is a tree for each gcd = 3 , ,
5. Notice also thatcontraction turns vertices of degree 3 into vertices of degree 4. The trees T and T have the same form; they are shown in Figure 6. The tree T is shown in Figure 7.The fundamental solutions are the solutions ( b, c ) for which b + c is minimal. Forgcd( b, c ) = 3 , ,
5, the fundamental solutions are (3 , , (4 , , (5 ,
5) respectively.Apart from the fundamental solutions, the vertices of T gcd all have degree 4. Thefundamental solutions of T and T have degree 2, while the fundamental solutionof T has degree 1. ¯ φ Figure 5.
Contraction of the ¯ φ edgesLet us look in more detail at the edges of the LEP solution trees T gcd . Considera LEP ( b, c ). As we saw in the previous section, k is determined by gcd( b, c ); k = 9 , , b, c ) = 3 , , bc = k ( m + n ) , b + c = kmn as before, and imposing n ≥ m , the values m, n are given by the following formulas: kn = bc + p b c − b + c ) km = bc − p b c − b + c ) . (12)For each i = 1 ,
2, let ϕ i denote the function induced by φ i on the set T gcd ofLEP pairs ( b, c ). Note that in the above notation, as b + c = kmn , we have( b, c ) = ( f sq, kmn − f sq ). From the definition of φ , the map ϕ leaves b and n unchanged and m is changed to m ′ = bn − m . Then under ϕ , the value of c ischanged to c ′ = km ′ n − b = kbn − knm − b = kbn − ( b + c ) − b = b c + b p b c − b + c ) − b − c. (from (11)) ∗ Thus ϕ : ( b, c ) b, b c + b p b c − b + c ) − b − c ! . Similarly, one finds that ϕ : ( b, c ) b, b c − b p b c − b + c ) − b − c ! . A calculation shows that ϕ ◦ ϕ = id. However it is not true that ϕ ◦ ϕ ( b, c ) =( b, c ) for all ( b, c ). In fact, the function ϕ is not injective and the function ϕ is not surjective; there is no solution ( b, c ) with b < c for which ϕ ( b, c ) is thefundamental solution.Similarly, analogous to ϕ , ϕ , interchanging the roles of b and c , we have twofurther maps: ψ : ( b, c ) bc + c p b c − b + c ) − c − b, c ! ,ψ : ( b, c ) bc − c p b c − b + c ) − c − b, c ! . After applying the maps ϕ i , ψ i , one reverses the image ( b ′ , c ′ ) if necessary so that b ′ ≤ c ′ . The edges in the trees T gcd are all obtained by applications of the fourmaps ϕ , ϕ , ψ , ψ .5. Recurrence relations in the branches of the LEP trees
The trees T , T , T contain a raft of interesting patterns that connect withphenomena that have been observed in other areas. We will give some simpleexamples. First we require a technical lemma concerning recurrence relations.Consider the real function f ( x ) = ux + v + p ( u − x + 2( u + 1) vx + w, where u, v, w are reals with u >
1. Notice that for large x the function f ( x ) isasymptotic to the linear function ( u + p ( u − x + v , which is monotonicallyincreasing as u >
1. If necessary, we restrict the domain of f ( x ) to a maximalhalf-infinite interval [ a, ∞ ) on which f ( x ) is defined, and has a well defined inverse. Lemma 5.
Consider the function f ( x ) = ux + v + p ( u − x + 2( u + 1) vx + w with the u > . Choose a in the domain of f ( x ) as described above and define a i recursively by a i = f ( a i − ) . Then a i satisfies the 3rd order recurrence relation a i = (2 u + 1) a i − − (2 u + 1) a i − + a i − . Remark . It is not difficult to see that if a function f ( x ) is of the form required byLemma 5, then for any affine change of variable, X = rx + s , the function inducedby f on X also has the same form, with the same value of u , while v changes to rv − us + s . In particular, as u >
1, one can choose s = rv/ ( u −
1) to make v become 0. ATTICE EQUABLE PARALLELOGRAMS 13 (3 , ,
5) (3 , , , ,
10) (3 , , , , , , , , , , , , , , , , , , , , , , , , Figure 6.
The two trees of LEPs with gcd( b, c ) = 3 and 5 respectively
Proof of Lemma 5.
The required recurrence relation is clearly invariant underaffine transformations. So, by the above remark, we may suppose that v = 0.Thus f ( x ) = ux + p ( u − x + w. We start by showing that f − ( x ) = ux − p ( u − x + w . Indeed, if f ( y ) = x ,then uy + p ( u − y + w = x . Rearranging and squaring gives y − uxy + x − w = 0, and solving for y gives y = ux ± p ( u − x + w . As u >
1, we have f ( x ) > x for sufficiently large x , so f − ( x ) < x and we take the negative root inthe formula for y . So f − ( x ) is as claimed.Let E := f ( f ( x )) − ((2 u + 1) f ( x ) − (2 u + 1) x + f − ( x )). We claim that E = 0.Indeed, let D = ( u − x + w and note that f ( f ( x )) = u x + u √ D + q ( u − ux + √ D ) + w and (2 u + 1) f ( x ) − (2 u + 1) x + f − ( x ) = (2 u − x + 2 u √ D. ∗ (4 ,
4) (4 ,
20) (4 , , , , , , Figure 7.
The tree of LEPs with gcd( b, c ) = 4So E simplifies to − [( u − x + u √ D ] + q ( u − ux + √ D ) + w. Note that as u >
1, we have ( u − x + u √ D >
0. Hence to show that E = 0 weneed to see that[( u − x + u √ D ] − [( u − ux + √ D ) + w ] = 0 . Expanding and simplifying the left-hand-side we are left with − ( u − x + D − w, which is 0, as claimed, by the definition of D .As E = 0 we have f ( f ( x )) = (2 u + 1) f ( x ) − (2 u + 1) x + f − ( x ), for all x in theappropriate domain. Replacing x by f ( x ) gives f ( f ( f ( x ))) = (2 u + 1) f ( f ( x )) − (2 u + 1) f ( x ) + x. Applying this equation to x = a gives the required recurrence relation. (cid:3) ATTICE EQUABLE PARALLELOGRAMS 15
For the function ϕ ( b, c ) = ( b, b c + b √ b c − b + c ) − b − c ) defined in the previoussection, let us fix b and look at the resulting function of c :(13) f : c b c + b p b c − b + c ) − b − c. One readily verifies that this function f has the form required by Lemma 5, with u = b − , v = − b, w = − b . Then Lemma 5 gives the recurrence relation(14) c i = ( b − c i − − ( b − c i − + c i − . Remark . While the function f of (13) is useful in its own right, we now brieflyoutline a proof of the recurrence relation (14) that doesn’t require Lemma 5. As wesaw in the previous Section, the map ϕ leaves b and n unchanged and m is changedto m ′ = bn − m . But this gives m ′ > n so we have to then interchange m ′ and n .Thus we can write n ′ = bn − m and m ′ = n . It follows that under two applications,we have m ′′ = n ′ = bn − m = bm ′ − m . Consequently, the value m satisfies thesecond order recurrence relation m i +1 = bm i − m i − . Similarly, n i +1 = bn i − n i − .According to a well known theorem of E. S. Selmer (see [2]) if one has a second orderrecurrence relation a i +1 = Aa i + Ba i − and x − Ax − B = ( x − α )( x − β ) with α = β ,then a i satisfies the third order recurrence relation a i +1 = Ca i + Da i − + Ea i − ,where x − Cx − Dx − E = ( x − α )( x − β )( x − αβ ). In particular, if a i +1 = ba i − a i − with b ≥
3, then a i +1 = ( b − a i − ( b − a i − + a i − . So the sequencesfor m and n both satisfy this third order recurrence relation, and hence so toodoes c = k ( m + n ) /b .We now examine the horizontal branches of the trees T , T , T that start at thefundamental solutions and head towards the right. These solutions have constant b and are defined by repeated applications of the map ϕ , or equivalently, repeatedapplications of the function f defined in (13). Example . For b = 3, with the above notation, (13) gives f : c c − r c − c − . Setting c = 6, corresponding to the fundamental LEP b = 3 , c = 6, the first fewterms of the sequence are 6 , , , , , u = , v = − , w = −
81. Thus c i = 8 c i − − c i − + c i − , for i ≥ . By Remark 6, the area of the LEP, 2( b + c ), is divisible by 9, and it is obviouslyeven. Let A = 2( b + c ) /
18 = (3 + c ) /
9. This is an affine transformation of c , so byRemark 7, the function induced by f on the variable A also has the form requiredby Lemma 5 with the same value of u . So by Lemma 5, A i = 8 A i − − A i − + A i − , for i ≥ . Setting A = 2, corresponding to the fundamental LEP b = 3 , c = 6, the first fewterms of the sequence are 1 , , , , ∗ Example . For b = 4, (13) gives f : c c − p c − c − . Setting c = 4, corresponding to the fundamental LEP b = 4 , c = 4, the first fewterms of the sequence are 4 , , , , , u = 7 , v = − , w = − c i = 15 c i − − c i − + c i − , for i ≥ . Dividing the c -values by 4 we have the sequence 1 , , , , , . . . . This is thesequence A103974 in OEIS [10]; it is the sequence of smaller sides a in ( a, a, a + 1)-integer triangles with integer area.Further, by Remark 6, the area, 2( b + c ) is divisible by 16. Let A = ( b + c ) / c ) /
8. As in the previous example, this is an affine transformation of c , so byRemark 7, the function induced by f on the variable A also has the form requiredby Lemma 5 with the same value of u . Thus A i = 15 A i − − A i − + A i − , for i ≥ . Setting A = 1, corresponding to the fundamental LEP b = 4 , c = 4, the first fewterms of the sequence are 1 , , , , Example . For b = 5, (13) gives f : c c −
10 + r · c − c − . Setting c = 5, corresponding to the fundamental LEP b = 5 , c = 5, the firstfew terms of the sequence are 5 , , , , , u = , v = − , w = − c i = 24 c i − − c i − + c i − , for i ≥ . The area 2( b + c ) is divisible by 5. Let A = 2( b + c ) /
5. We employ Lemma 5 withthe same value of u . Thus A i = 24 A i − − A i − + A i − , for i ≥ . Setting A = 4, corresponding to the fundamental LEP b = 5 , c = 5, the first fewterms of the sequence are 4 , , , , Diagonals, heights and altitudes
Let us first fix some terminology and notation; see Figure 8.
Definition 2.
Consider a non-square LEP P . We denote the length of its long(resp. short) diagonal d l (resp. d s ). The heights of P are the distances betweenopposite sides; we denote the long (resp. short) height h l (resp. h s ). Each diagonal d partitions P into two congruent triangles T . We will call the distance from d to ATTICE EQUABLE PARALLELOGRAMS 17 the third vertex of T an altitude of P . We call the altitude from d s (resp. d l ) the long (resp. short ) altitude and denote it η l (resp. η s ). Remark . Our notion of altitude is not universal. Some authors use the termaltitude for the concept we have called height. b c h l h s d l d s η l η s Figure 8.
Diagonals, heights and altitudes
Lemma 6.
Suppose a LEP P has sides b, c with b ≤ c . Then (a) h l = 2( b + c ) b , h s = 2( b + c ) c . (b) η l = 2( b + c ) d s , η s = 2( b + c ) d l .Proof. By equability, the area of P is 2( b + c ), but the area is obviously also bh l and ch s . This gives (a). But the area of P is also twice the area of thetriangle determined by each diagonal. So the area of P is both η l d s and η s d l . Thisgives (b). (cid:3) Remark . If h s is an integer, then h s = 2 + bc by Lemma 6(a), and so either c = b or c = 2 b . These cases were treated in Corollary 1. Note also that the abovelemma also gives ( h s − h l −
2) = 4. So h s , h l are both integers only in the cases h s = h l = 4 and h s = 3 , h l = 6. The first is the 4 × × Theorem 5.
For every LEP the altitudes satisfy < η l ≤ √ and < η s ≤ √ .Proof. Suppose a LEP P has sides b, c with b ≤ c . With the notation used inSection 3, we have bc = k ( m + n ) , b + c = kmn, p b c − b + c ) = k ( n − m ),for some positive integer k and relatively prime integers m, n with m ≤ n . From ∗ Lemma 2, the diagonals are given by d l = ( b + c ) + p b c − b + c ) = k m n − k ( m + n ) + 2 k ( n − m ) = m ( k n − k ) ,d s = ( b + c ) − p b c − b + c ) = k m n − k ( m + n ) − k ( n − m ) = n ( k m − k ) . The long altitude is η l = 2( b + c ) d s = 2 knmn √ k m − k . Squaring and rearranging gives km = η l η l − . As m ≥ k ≥ η l η l − ≥ ≥ η l , that is η l ≤ √
5. Furthermore, η l = 2 knmn √ k m − k > knmn √ k m = 2 . Arguing in the same manner one finds η s > kn = η s η s − . As we saw inSection 4, for k = 9 (resp. 8, resp. 5), the fundamental solution has ( m, n ) = (1 , kn is 8. Rearranging 8 ≤ η s η s − gives η s ≤
8, that is η s ≤ √ (cid:3) Remark . The bound η l = 2 √ η s = 2 √ × Proposition 3.
For every LEP, the two diagonals and the two altitudes are irra-tional.Proof.
As we saw in the above proof, d l = m ( k n − k ). So the longer diagonalhas integer length only when k n − k is a square. But k = 5 , k = 5 (resp. 8), the expression k n − k is divisible by 5 (resp. 4 k = 2 ) butnot by 5 (resp. 2 ) and is hence never a square. For k = 9, one has k n − k =3 (9 n − n − n . Thus the long diagonalis never an integer, and hence by Remark 1, the long diagonal is irrational forevery LEP. Similarly, the short diagonal is always irrational as k m − k is nevera square. Consequently, by Lemma 6(b), the altitudes η l , η s are also irrational forevery LEP. (cid:3) The following immediate corollaries of Proposition 3 are each generalisations ofProposition 2.
Corollary 2.
No LEP can be partitioned into the union of two Heronian triangles.
Corollary 3.
There exists no LEP having a pair of opposite vertices with thesame x -coordinate, or the same y -coordinate. ATTICE EQUABLE PARALLELOGRAMS 19
Figure 9.
A parallelogram that isn’t a LEP
Proposition 4.
If a LEP P contains the origin O and vertices O, A, B, C incyclic order with long diagonal OB = d l belonging to the 1st quadrant, then allfour vertices belong to the first quadrant. d l η s O A ( x, y ) BC Figure 10.
A LEP configuration that can’t exist
Proof.
By applying if necessary a symmetry along the x = y axis we suppose that A lies in the 4th quadrant as in Figure 10. Since the diagonals are irrational thevertex B can’t be on the x -axis. Moreover, as P has an acute angle at O , thefoot of the altitude from A to d l also lies in the first quadrant. Hence, letting A = ( x, y ), the distance from A to d l is greater than | y | . Thus, by Theorem 5, | y | < η s ≤ √
2. Hence, as y is an integer, | y | is 1 or 2. Let OA have length b .Then b is x + 1 or x + 2. But this is impossible for x > (cid:3) Corollary 4.
Every LEP is congruent to a LEP in the 1st quadrant with a vertexat the origin. Pythagorean Equable Parallelograms
In general there are two ways in which parallelograms can be circumscribedby a rectangle so that the rectangle and parallelogram share a common diagonal;the rectangle may have sides extending the long sides of the parallelogram, or theshort sides of the parallelogram; see Figure 11. ∗ Figure 11.
Rectangles circumscribing a parallelogram
Definition 3.
A LEP is said to be
Pythagorean if it is circumscribed by a rectanglehaving integer side lengths so that the rectangle and parallelogram share a commondiagonal as in Figure 11.This terminology is justified by the equivalent condition (b) in the followingresult.
Proposition 5.
Consider a LEP P having sides b, c with b ≤ c . The followingconditions are equivalent: (a) P is Pythagorean, (b) P is circumscribed by a rectangle R such that the complement of P in R is the union of two Pythagorean triangles, (c) b divides c , (d) P can be drawn as a LEP with a horizontal pairs of sides. h l a cb Figure 12.
Integer h l = ⇒ integer a Proof of Proposition 5.
Suppose that P is a LEP with vertices O, A, B, C , in cyclicorder, and that P has sides b, c with b ≤ c .(a) = ⇒ (d). Suppose that P is Pythagorean and that the distance betweensides OA and BC is an integer, h say. Rotate and translate P so that it has the ATTICE EQUABLE PARALLELOGRAMS 21 side OA along the positive x -axis, with the vertex O at the origin. We want toverify that moved in this way, P is still a LEP. Since P has integer side lengths, A has integer coordinates. By assumption, B, C have integer y -coordinate, equalto h . From the Pythagorean hypothesis, B has integer x -coordinate, z say. Then x -coordinate of C is z minus the x -coordinate of A . So P is a LEP.(d) = ⇒ (c). If a LEP P has a horizontal pairs of sides, then the fact thatthe coordinates of the vertices are integers implies that the distance h between thehorizontal sides is an integer. So either h l or h s is an integer. If h s is an integer,then as h s = 2 + bc by Lemma 6(a), and b ≤ c , we have either c = b or c = 2 b . Ineither case, by Lemma 6(a) again, h l = 2 + cb is also an integer. Thus b divides2 c .(c) = ⇒ (b). Since h l = 2 + cb , hence h l is an integer. Consider the rectangle R that circumscribes P , sharing a common diagonal with P and having sides thatare extensions of the sides of P of length b , as in Figure 12 (but not necessarilywith the sides b horizontal). So R has sides h l and b + a , for some a , and we arerequired to show that a is an integer. By Lemma 6(a), we have a = c − h l = c − b + c ) b = b c − b + c ) b , so a = √ b c − b + c ) b , which is a rational by Theorem 1. Since h l = 2 + cb is aninteger, 2 c = bi for some integer i . Then a = b c − b + c ) b = c − (2 + i ) , which is an integer. So a is rational and the square root of an integer. Hence a isan integer.(b) = ⇒ (a). This is immediate. (cid:3) Remark . The smallest non-Pythagorean LEP has area 180 and sides 25 and 65;it can be constructed in the first quadrant with vertices (0 , , (25 , , (32 , , (7 , Proof of Theorem 3.
Consider a Pythagorean LEP P with sides b, c . By Propo-sition 5, b divides 2 c , and by Theorem 2, gcd( b, c ) = 3 , b is odd, then b divides c , so b = gcd( b, c ) and b is 3 or 5. If b is even, say b = 2 b ′ , then b ′ divides c so b ′ divides gcd( b, c ). Thus b ′ is 2 , , b is 4 , , b, c ) = 4 neither b nor c is divisible by 8. So b is 3,4,5,6 or 10, as claim in Theorem 3. In remains to exhibit a family of suchPythagorean LEPs in each case.In the notation of previous sections, bc = k ( m + n ) , b + c = kmn for positiveintegers k, m, n where m, n are relatively prime with m < n . For convenience, letus restate (8): km + kn + b = kbmn. Moreover, by Lemma 3, k is either 5, 8 or 9, corresponding to gcd( b, c ) = 3 , , b = 3 we have k = 9 and (8) gives m + n + 1 = 3 mn . Setting x = n − m, y = m + n we obtain the Pell-like equation y − x = 4. Conversely,note that if ( x, y ) is a solution to this equation, then x, y necessarily have the same ∗ parity and we can set m = y − x , n = y + x to obtain an integer solution ( m, n ) to m + n + 1 = 3 mn . Then c = kb ( m + n ) = x + y )2 .For b = 4 we have k = 8 and (8) gives m + n + 2 = 4 mn . Note then that m , n must have same parity. Setting 2 x = n − m , 2 y = n + m we obtain thePell equation y − x = 1. Conversely, note that if ( x, y ) is a solution to thisequation, then we can set m = y − x, n = y + x to obtain an integer solution ( m, n )to m + n + 2 = 4 mn . Then c = kb ( m + n ) = 4( x + y ).For b = 5 we have k = 5 and (8) gives m + n + 5 = 5 mn . Setting x = n − m, y = m + n we obtain the Pell-like equation 3 y − x = 20. Conversely,note that if ( x, y ) is a solution to this equation, then x, y necessarily have the sameparity and we can set m = y − x , n = y + x to obtain an integer solution ( m, n ) to m + n + 5 = 5 mn . Then c = kb ( m + n ) = x + y .Similarly, for b = 6 (resp. b = 10), we have k = 9 (resp. k = 5) and (8) gives m + n + 4 = 6 mn (resp. m + n + 20 = 10 mn ). Note that m, n necessarily havethe same parity and we can set x = n − m , y = m + n . This gives the Pell (resp. Pell-like) equation y − x = 1 (resp. 2 y − x = 5). Conversely, note that if ( x, y )is a solution to this equation, then we can set m = y − x, n = y + x to obtainan integer solution ( m, n ) to m + n + 4 = 6 mn (resp. m + n + 20 = 10 mn ).Using c = kb ( m + n ), for b = 6 we have c = 3( x + y ) and for b = 10 we have c = x + y . (cid:3) We complete this section by connecting the above solution families F1-F5 to thematerial in the earlier Sections. In particular, we locate the Pythagorean equableparallelograms in the trees of LEPs of Figure 6 and 7.
F1: b = 3 . The solutions ( x, y ) to the Pell-like equation y − x = 4 are wellknown to be ( L i , F i ), starting at ( F , L ) = (0 , , ( F , L ) = (1 , F stands for the Fibonacci numbers and L for the Lucas numbers. Both L i and F i satisfy the recurrence relation a i = 3 a i − − a i − , with different initial conditions.So the first few solutions of equation y − x = 4 are: (0 , , (1 , , (3 , , (8 , , c -values, given by c = x + y )2 , are: 6, 15, 87, 582, 3975.We saw the corresponding LEPs (3 , c ) previously in Example 1; in Figure 6 theylie on the central horizontal branch to the right of the fundamental solution (3 , c -values are generated recursively by the formula f ( c ) = c − q c − c −
81, or alternatively, using the recurrence relation c i = 8 c i − − c i − + c i − . Notice that by Theorem 1, these c values are precisely those numbers c = 3 i for which 5 i − i − F2: b = 4 . The solutions ( x, y ) to Pell’s equation y − x = 1 are wellknown to be given by the recurrence relation a i = 4 a i − − a i − . See OEIS en-tries A001075 and A001353 [10]. The first few solutions of y − x = 1 are:(0 , , (1 , , (4 , , (15 , , (56 , c -values, given by c = 4( x + y ),are: 4, 20, 260, 3604, 50180. We saw the corresponding LEPs (4 , c ) in Example 2;in Figure 7 they lie on the central horizontal branch to the right of the fundamentalsolution (4 , c -values can be generated recursively by the formula f ( c ) =7 c − √ c − c −
16, or by the recurrence relation c i = 15 c i − − c i − + c i − . ATTICE EQUABLE PARALLELOGRAMS 23
Notice that by Theorem 1, these c values are precisely those numbers c = 4 i forwhich 3 i − i − F3: b = 5 . The Pell-like equation 3 y − x = 20 is less common. Its solutions( x, y ) are given by the recurrence relation a i = 5 a i − − a i − . The first few so-lutions are: (1 , , (2 , , (7 , , (11 , , (34 , , (53 , , (163 , c -values, given by c = x + y , are: 5, 10, 85, 205, 1930, 4685, 44285. The positionof the corresponding LEPs (5 , c ) is more complicated than what we saw for b = 3and b = 4. Every second solution, starting at (5 , , c -values can be generated recursively by the formula f ( c ) = 232 c −
10 + r · c − c − , or by the recurrence relation c i = 24 c i − − c i − + c i − . The other solutionsare on the central horizontal branch of Figure 6 to the left of the fundamentalsolution (5 , c -values are generated recursively by the same formula (andrecurrence relation) but starting at c = 10.Note that in the proof of Theorem 3 the equation 3 y − x = 20 was derivedfrom m + n + 5 = 5 mn . This latter equation is well known; indeed, it is easyto see that for the solutions m, n , the first component comprise those numbers m for which 21 m −
20 is a square. See entry A237254 in [10]. By Theorem 1, the c values for b = 5 are precisely those numbers c = 5 i for which 21 i − i − F4: b = 6 . The solutions ( x, y ) to Pell’s equation y − x = 1 are well knownto be given by the recurrence relation a i = 6 a i − − a i − . See OEIS entries A001541,A001542 [10]. The first few solutions are: (0 , , (2 , , (12 , , (70 , , (408 , c -values, given by c = 3( x + y ), are: 3, 39, 1299, 44103, 1498179.The corresponding LEPs (6 , c ) occur on the central horizontal branch of Figure6, to the left of the fundamental solution (3 , c -values of thesolutions on this branch we can use Equation 13 with b = 6. These c -values canbe generated recursively by the formula f ( c ) = 17 c −
12 + 6 p c − (6 + c ) , and by the recurrence relation c i = 35 c i − − c i − + c i − . By Theorem 1, these c values are precisely those numbers c = 3 i for which 2 i − i − F5: b = 10 . Like the b = 5 case, the Pell-like equation 2 y − x = 5 is notvery common. Its solutions ( x, y ) are given by the recurrence relation a i = 10 a i − − a i − . The first few solutions are: (1 , , (3 , , (13 , , (31 , , (129 , , (307 , c -values, given by c = x + y , are: 5, 25, 425, 2405, 41605, 235625.The corresponding LEPs (10 , c ) occur on the first vertical branch to the left of thefundamental solution (5 ,
5) in Figure 6. Every second solution, starting at (10 , c -values we can useEquation 13 with b = 10. This gives the equation f ( c ) = 49 c −
20 + 10 p c − (10 + c ) , ∗ and by (14), the recurrence relation c i = 99 c i − − c i − + c i − . The other solutionsappears below the central horizontal axis. Their c -values are generated recursivelyby the same formula (and recurrence relation) but starting at c = 25.Note that in the proof of Theorem 3 the equation 2 y − x = 5 was derivedfrom m + n + 20 = 10 mn . This latter equation is well known; indeed, it is easyto see that for the solutions m, n , the first component comprise those numbers m for which 6 m − c values for b = 10 are precisely those numbers c = 5 i for which 6 i − i − Remark . The Pell-like equations in families F1 – F5 can all be put in thefollowing form:(15) ay − ( a + 4) x = d, where a, d are integers and d is even if a is odd. Table 1 gives the values of a, d for the five families.A direct calculation shows that if ( x, y ) is a solution to (15), then anothersolution is given by (cid:18) x ′ y ′ (cid:19) = 12 (cid:18) a + 2 aa + 4 a + 2 (cid:19) (cid:18) xy (cid:19) = x + a ( x + y )2 x + y + a ( x + y )2 ! . This was proved by R´ealis for d = ± a is odd, then from (15), as d is even, the numbers x, y must have the same parity, so x ′ , y ′ are integers). Furthermore, provided (15) hasone solution, ( x , y ) say, this process gives a sequence of solutions ( x i , y i ), and itis easy to verify that it satisfies the following recurrence relation:( x i , y i ) = ( a + 2)( x i − , y i − ) − ( x i − , y i − ) . Note that (15) does not have a solution for all a and d . For example, for a =3 , d = 4, the equation is 3 y − x = 4. Modulo 3 this is − x ≡
1, which has nosolution. Family F1 F2 F3 F4 F5 a d Table 1.Appendix: The Equable Triangles Theorem
The Equable Triangles Theorem says that there are only 5 equable triangles.For the early history of the theorem, Dickson [3, pp. 195, 199] cites Whitworth andBriddle in Math. Quest. Educational Times 5, 1904, 54–56, 62–63, but we havebeen unable to locate this/these works. Equable triangles have been investigatedin several works [5, 6, 11, 13]. However the only proof of the Equable TrianglesTheorem we have seen in the literature is by Arthur H. Foss [4]. In this appendixwe supply a different proof.
ATTICE EQUABLE PARALLELOGRAMS 25 b = k k ka c B C AD
Figure 13.
Equable trianglesFor equable triangles with sides a, b, c , Heron’s formula gives( a + b + c )( − a + b + c )( a − b + c )( a + b − c ) = 16( a + b + c ) . Let u = − a + b + c, v = a − b + c, w = a + b − c , so that a = v + w , b = u + w , c = u + v .Then our equation is(16) uvw = 16( u + v + w ) , and we look for solutions u, v, w , all of the same parity. Further, we may assume u ≤ v ≤ w . Note that, since u, v, w , have the same parity, so from (16), u, v, w arenecessarily even. Let u = 2 x, v = 2 y, w = 2 z , so a = y + z, b = x + z, c = x + y .Then xyz = 4( x + y + z ). Thus y ≤ z = 4( x + y ) xy − , so xy − y − x ≤
0. Hence x ≤ y ≤ √ x x so x ≤ √ x . Hence ( x − ≤ x ). Thus x − x ≤
0, whichgives x ≤
3. Then y ≤ √ x x ≤ √ , since the function √ x x is decreasing for positive x . So, as y is an integer, y ≤
8. Then, considering the values x ≤ y ≤ z = x + y ) xy − , we find thefollowing integer values for x, y, z :1 , ,
24 1 , ,
14 1 , , , ,
10 2 , , , which give the values for a, b, c :6 , ,
29 7 , ,
20 9 , ,
17 5 , ,
13 6 , , . This completes the proof. Note that the last two are Pythagorean triples, whileintriguingly, for the other three cases, the corresponding triangle is the complementof a Pythagorean triangle of the form 3 k, k, k in a larger Pythagorean triangle;see Figure 13. The three cases correspond to the values k = 2 , , ∗ References
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