On primitive Pythagorean triples of special forms
aa r X i v : . [ m a t h . N T ] F e b ON PRIMITIVE PYTHAGOREAN TRIPLES OF SPECIALFORMS
ANDREW SCHMELZER AND SUNIL CHETTY
Abstract.
We explore primitive Pythagorean triples of special forms ( a, b, b + g ) and ( a, a + f, c ), with g, f ∈ Z + . For each g and f , we provide a method togenerate infinitely many such primitive triples. Lastly, for each g , we describethe asymptotic density of primitive ( a, b, b + g ) triples within all primitivetriples of the same parity. Introduction A Pythagorean Triple (PT) is a triple of integers ( a, b, c ) ∈ Z such that a + b = c . A Pythagorean triple is a primitive Pythagorean triple (PPT) if d > d divides a, b, and c implies d = 1. To see a Pythagorean triple is primitive, it sufficesto show d > a , b , or c implies d = 1.The two theorems below are classical results, and we include a proof of the secondfor completeness sake and to demonstrate the spirit of several arguments to followin § Theorem 1.1.
All Pythagorean triples ( a, b, c ) take the form a = r − s , b = 2 rs , c = r + s (up to switching the parity for a and b ) for some r, s ∈ Z with s < r .Proof. See, for example, [2, § VII.2] (cid:3)
Theorem 1.2.
In the notation of Theorem 1.1, a Pythagorean triple is primitiveif and only if gcd( r, s ) = 1 .Proof.
If ( a, b, c ) is a PPT and gcd( r, s ) = d then d is also a factor of a = r − s and b = 2 rs , so d = 1.Now suppose gcd( r, s ) = 1. The parity of r and s must be opposite, otherwise a and b would both be even. So we take r odd and s even. If d is a common factorof a = r − s , b = 2 rs , and c = r + s , then d is odd because a is odd. Also, d divides the sum and difference of c and a , i.e. 2 s = c − a and 2 r = c + a , andhence d divides gcd( r, s ), so d | (cid:3) Because there are infinitely many pairs r, s ∈ Z with gcd( r, s ) = 1, it followsfrom Theorem 1.2 that there are infinitely many primitive Pythagorean triples.However, the result does not immediately provide information about families ofprimitive triples that share a particular quality. Date : February 10, 2021.2010
Mathematics Subject Classification.
Key words and phrases. primitive Pythagorean triples, Pell equation, density.We thank the College of Saint Benedict and Saint John’s University for supporting this researchduring Summer 2019. We thank Brandon Alberts for his helpful conversation ans suggestionsregarding the density results. Computational exploration that guided this research was conductedusing SageMath [6]. Primitive triples of the form ( a, b, b + g )Using only elementary algebra, one can show that there are infinitely manytriples of the form ( a, b, b + 1). Extending this, one can ask if there are infinitelymany PPT of the form ( a, b, b + g ), with g > Conditions on g .Lemma 2.1. In a PPT, exactly one of a, b is odd, and c is always odd.Proof. If both a and b are even, then c is also even and ( a, b, c ) is not primitive. If a and b are both odd, then a ≡ b ≡ c ≡ a, b is odd. (cid:3) The following proposition gives the form of g in primitive ( a, b, b + g ) triples. Proposition 2.2. If ( a, b, b + g ) is a PPT then g = m , with m odd, or g = 2 m ,with m any positive integer.Proof. Let ( a, b, b + g ) be a PPT, and suppose g is odd. Because of Lemma 2.1, itmust be b = 2 rs , otherwise c = b + g would be even and the triple would not beprimitive. Then by Theorem 1.1 our triple fits the form a = r − s , b = 2 rs , and b + g = r + s . Solving for g yields g = r + s − rs = ( r − s ) , so for m = r − s we have g = m .Suppose m is even. Note that a = ( r − s )( r + s ). If r − s is even, then r + s is even too, and so a would be even. Then both of a, b would be even, and thatcontradicts Lemma 2.1, so m must be odd.Now, suppose g is even. Then, again using Theorem 1.1, our triple fits the form a = 2 rs , b = r − s , and b + g = r + s . As before, we use Lemma 2.1 to know b = r − s , otherwise b and c = b + g are even. Solving for g yields g = 2 s , so for m = s and we have g = 2 m . (cid:3) Generating primitive ( a, b, b + g ) triples. Now that we know the shape ofall possible g , we discuss generating the triples. Lemma 2.3.
Any PPT of the form ( a, b, b + g ) is determined by a and g .Proof. We know that a + b = ( b + g ) = b + 2 bg + g . Solving for b yields b = a − g g . (cid:3) Theorem 2.4.
For each g ∈ { (2 k + 1) : k ∈ Z } ∪ { m ) : m ∈ Z } , there are aninfinite number of PPTs of the form ( a, b, b + g ) .Proof. For each case of g , we choose values t and q , and construct a family ( a n ) =( tn + q ) parametrized by n ∈ Z + in order to define an infinitely family of PT( a n , b n , b n + g ).Let g = m for some odd m . Let t = 2 m , q = m , so a n = 2 mn + m = m (2 n + 1).Then, applying Lemma 2.3, b n = m (2 n + 1) − m m = (2 n + 1) − m n + 1 − m )(2 n + 1 + m )2 . N PRIMITIVE PYTHAGOREAN TRIPLES OF SPECIAL FORMS 3
Note that in the right-most expression both elements of the numerator are even,making b n even. Then our triple is (cid:18) m (2 n + 1) , (2 n + 1) − m , (2 n + 1) + m (cid:19) . Arguing as in Theorem 1.2, this form generates a PPT whenever gcd( m, n +1) = 1There are an infinite number of primes of the form 2 n + 1 which do not divide m ,so that condition holds an infinite number of times.Now, let g = 2 m for some odd m . Let t = 2 m , q = 0, so a n = tn = 2 mn , andthen b = a − g g = 4 m n − m m = n − m . Then we have (cid:0) mn, n − m , n + m (cid:1) is a PPT whenever gcd( n, m ) = 1, exactlyby Theorem 1.2. There are an infinite number of primes n not dividing m , so thatcondition holds an infinite number of times.Next, let g = 2 m for some even m . Let t = 4 m and q = 2 m , so a n = tn + q =2 m (2 n + 1) . Then b n = a − g g = 4 m (2 n + 1) − m m = (2 n + 1) − m . Then we have (cid:0) m (2 n + 1) , (2 n + 1) − m , (2 n + 1) + m (cid:1) . This form generates a PPT whenever gcd(2 n + 1 , m ) = 1, by Theorem 1.2 with r = 2 n + 1 and s = m . There are an infinite number of primes of the form 2 n + 1not dividing m , so that condition holds an infinite number of times.Thus, for every g ∈ { (2 m + 1) : m ∈ Z } ∪ { m ) : m ∈ Z } , there are an infinitenumber of PPTs of the form ( a, b, b + g ). (cid:3) Note the above construction of ( a n , b n , b n + g ) provides a family of PT withstrictly increasing sequence of first coordinates ( a n ) g , for each g as in Theorem 2.4.In turn, we have a subfamily of PPTs with strictly increasing first coordinates. Corollary 2.5.
Every PPT of the form ( a, b, b + g ) , is contained in one of thefamilies ( a n , b n , b n + g ) . Moreover, ( a, b, b + g ) is obtained from r n and s n (as inTheorem 1.2) according to: g = m g = 2 m g = 2 m m odd m odd m even r n = (2 n + 1 + m ) r n = n r n = 2 n + 1 s n = (2 n + 1 − m ) s n = m s n = m gcd(2 n + 1 , m ) = 1 gcd( n, m ) = 1 gcd(2 n + 1 , m ) = 1 Proof.
For g odd, we have m odd. The proof of Proposition 2.2 shows m = r − s , a = ( r − s )( r + s ), and both are odd. So a = mk for some odd k . For g even,Proposition 2.2 shows m = s and a = 2 rs , so a = mk , and both cases of k evenand k odd arise as a n . (cid:3) ANDREW SCHMELZER AND SUNIL CHETTY Primitive triples of the form ( a, a + f, c )We now discuss a second form of PPT, with a fixed difference f = b − a . First,we look into the form of f .3.1. Conditions on f .Proposition 3.1. Suppose p is a prime dividing f . If ( a, a + f, c ) is a PPT, then p ≡ ± (mod ).Proof. Let p be a prime dividing f and consider the defining equation modulo p ,i.e. c ≡ a + ( a + f ) ≡ a mod p. Notice that p does not divide a , otherwise p also divides b and c , and so ( a, a + f, c )is not primitive. Now, ( ca − ) ≡ p , making 2 a quadratic residue, and byquadratic reciprocity (see [2, § III.5] or [4, § p ≡ ± (cid:3) Before continuing, we recast the defining equation for the PPT in question. Let( a, a + f, c ) be a PPT. Then, by elementary algebra, we see a + ( a + f ) = c ⇔ a + 2 af + f = c ⇔ a + 4 af + 2 f − c = 0 ⇔ (2 a + f ) + f − c = 0 ⇔ (2 a + f ) − c = − f , with the last in the form of a Pell equation x − y = − f .Pell equations are well-studied Diophantine equations, and expanding our viewto the quadratic ring of integers Z [ √
2] is a particularly fruitful strategy. We nowrecord some useful information about Z [ √ N on Z [ √
2] by N ( β ) = β ¯ β . Theorem 3.2. Z [ √ is a Euclidean domain. More specifically, if α, β ∈ Z [ √ and β = 0 then there exist unique γ, ρ ∈ Z [ √ such that α = βγ + ρ with | N ( ρ ) | < | N ( β ) | .Proof. We employ a classical argument for quadratic rings, as in [4, § Z [ i ].Let α ∈ Z [ √
2] and 0 = β ∈ Z [ √ αβ = α ¯ ββ ¯ β , and both α ¯ β, β ¯ β ∈ Z [ √ αβ ∈ Q ( √ αβ = r + s √
2, with r, s ∈ Q . Let u, v ∈ Z such that | r − u | ≤ and | s − v | ≤ , and define γ = u + v √ ρ = α − βγ . By the multiplicativityof the the norm map N , | N ( ρ ) | = (cid:12)(cid:12)(cid:12)(cid:12) N (cid:18) αβ − γ (cid:19) N ( β ) (cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12) N (cid:18) αβ − γ (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) | N ( β ) | with (cid:12)(cid:12)(cid:12)(cid:12) N (cid:18) αβ − γ (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12) ( r − u ) − s − v ) (cid:12)(cid:12) . Since | r − s | ≤ and | s − v | ≤ , we have 0 ≤ ( r − u ) ≤ and 0 ≤ ( s − v ) ≤ ,and in turn (cid:12)(cid:12)(cid:12) N (cid:16) αβ − γ (cid:17)(cid:12)(cid:12)(cid:12) <
1. We conclude | N ( ρ ) | < | N ( β ) | as desired.The uniqueness of γ and ρ follow from the condition that | r − u | , | s − v | ≤ . (cid:3) Corollary 3.3. If p ⊂ Z [ √ is an ideal, then p = ( α ) for some α ∈ Z [ √ .Moreover, every non-unit α ∈ Z [ √ factors into primes uniquely. N PRIMITIVE PYTHAGOREAN TRIPLES OF SPECIAL FORMS 5
Proof.
Both parts of the claim are applications of classical results (e.g. [3, Theorems18.3-18.4]) in the theory of rings, following Theorem 3.2. (cid:3)
In addition to the above broad structural results, we recall the following well-known factorization of particular ideals in Z [ √ Theorem 3.4. If p ∈ Z is a prime such that p ≡ ± , then ( p ) = p ¯ p , with p , ¯ p prime ideals in Z [ √ .Proof. See, for example, [4, Theorem 13.1.3]. (cid:3)
Generating ( a, a + f, c ) triples. Let γ = 1 + √ , and δ = γ = 3 + 2 √ Theorem 3.5.
All solutions of x − y = − are given by ± γδ m , for m ∈ Z .Proof. See, for example, [5, § (cid:3) We see a recurrence within solutions of the classical negative Pell equation inthe previous theorem.
Proposition 3.6.
Suppose δ = 3 + 2 √ is as above, and t = j + k √ . Define, for n ≥ , sequences A n and B n by tδ n = A n t + B n (2 k + j √ . Then A = 1 , A = 3 , and A n = 6 A n − − A n − , for n ≥ , B = 0 , B = 2 , and B n = 6 B n − − B n − , for n ≥ .Proof. We first establish the initial conditions and then intermediate recurrencerelations for A n and B n . The case n = 0 is clear as tδ = t . Next, tδ = 3( j + k √
2) + 2 √ j + k √
2) = 3 t + 2(2 k + j √ , showing that A = 3 and B = 2.Now, assume A k and B k satisfy the given defining equation for all k < n . Wecompute tδ n = (cid:0) A n − ( j + k √
2) + B n − (2 k + j √ (cid:1) (3 + 2 √ A n − j + 3 A n − k √ A n − k + 4 B n − j + 6 B n − k +2 A n − j √ B n − j √ B n − k √
2= (3 A n − + 4 B n − ) j + (3 A n − + 4 B n − ) k √ A n − + 6 B n − ) k + (2 A n − + 3 B n − ) j √
2= (3 A n − + 4 B n − ) t + (2 A n − + 3 B n − )(2 k + j √ . Thus, we see A n = 3 A n − + 4 B n − and B n = 2 A n − + 3 B n − . To finish, we showthat these same mixed recurrence relations are satisfied when A n = 6 A n − − A n − and B n = 6 B n − − B n − .Consider x n and y n defined by the two recurrence relations of the conclusion ofthe claim. Suppose the mixed recurrence relations on A n and B n from the previousparagraph hold for x k and y k , for every k < n . We compute x n = 6(3 x n − + 4 y n − ) − x n − y n = 6(2 x n − + 3 y n − ) − y n − = 3 x n − + 3(3 x n − + 4 y n − ) − x n − = 3 y n − + 3(2 x n − + 3 y n − ) − y n − = 3 x n − + 8 x n − + 12 y n − = 3 y n − + 8 y n − + 6 x n − = 3 x n − + 4(2 x n − + 3 y n − ) = 3 x n − + 2(3 x n − + 4 y n − )= 3 x n − + 4 y n − = 3 y n − + 2 x n − . ANDREW SCHMELZER AND SUNIL CHETTY
Thus, when satisfying the same initial conditions, we see A n = x n and B n = y n ,for all n ≥ (cid:3) Let ( a, a + f, c ) be a PPT. Recall that a + ( a + f ) = c if and only if (2 a + f ) − c = − f . Now suppose f = p t . . . p t n n , with p i ∈ Z , t i ∈ Z + . By Proposition 3.1,we know that all the primes p i dividing f satisfy p i ≡ ± p i ) splits over Z [ √
2] according to Theorem 3.4, we know that the ideal ( f ) canalso be split into conjugate pair factors over Z [ √ f ) = p t ¯ p t . . . p t n n ¯ p t n n and C f := (Y i q t i i : q i = p i or q i = ¯ p i , for each 1 ≤ i ≤ n ) . Note that if u ∈ C f , then whenever q i | u , we have p i ∤ u . Also u ∈ C f implies u ¯ u = f , and hence N ( u ) = f . Theorem 3.7.
Recall γ = 1 + √ , δ = γ ∈ Z [ √ . All the PPTs of the form ( a, a + f, c ) are also of the form a = X − f , b = X + f , and c = Y , where X, Y arethe components of ± γδ m u , for some m ∈ Z and some u ∈ C f .Proof. Let ( a, a + f, c ) be a PPT, so (2 a + f ) − c = − f . Let u ∈ C f , so we have((2 a + f ) − √ c )((2 a + f ) + √ c ) = − u ¯ u . Without loss of generality, let p i | u and ¯ p i | ¯ u . Suppose that both p i and ¯ p i divide(2 a + f + √ c ). Then p i ¯ p i | (2 a + f + √ c ), i.e. p i | (2 a + f + √ c ), and since p i ∈ Z , we know p i divides c .We also know that p i | f , so p i divides two terms of (2 a + f + √ c ), hence alsomust divide 2 a . From p i | f , we also know p i ≡ ± p i does notdivide 2. So p i | a .Now, we have that p i | a, a + f, c , contradicting that ( a, a + f, c ) is a PPT. Thus,for any prime p i dividing (2 a + f + √ c ) , we know that ¯ p i ∤ (2 a + f + √ c ). Sinceno prime and it’s conjugate divide u , it must be that u | (2 a + f + √ c ).Let α = 2 a + f + √ c . From N ( u ) = f and N ( α ) = − f , we now have N (cid:0) α u (cid:1) = −
1. Then, by Theorem 3.5, the components x + √ y of α u are solutionsto x − y = −
1. So α u = ± γδ m , for some m . Then α = 2 a + f + √ c = ± γδ m u , and in turn X = 2 a + f, Y = c. So a = X − f , c = Y . (cid:3) The above theorem allows us to produce infinitely many triples of the desiredform from solutions to the Pell equation x − y = −
1. We now restrict to thecase f ≡ ± Algorithm 3.8.
Recall γ = 1 + √ , δ = γ ∈ Z [ √ f with f ≡ ± a, a + f, c ) can be generated as follows:(1) Factor the ideal ( f ) over Z [ √
2] as ( f ) = p ¯ p .(2) Determine a generator u ∈ Z [ √
2] of p . The restriction on f is for convenience and clarity of notation. Both Algorithm 3.8 and Theorem3.9 can be generalized to any f satisfying the conditions of § Each step can be accomplished with functionality available in Sage [6]
N PRIMITIVE PYTHAGOREAN TRIPLES OF SPECIAL FORMS 7 (3) Choose m ∈ Z + . Set A = 1, A = 3, B = 0, and B = 2.For 2 ≤ i ≤ m , compute A i = 6 A i − − A i − and B i = 6 B i − − B i − (4) Set X + Y √ A m + 2 B m ) + ( A m + B m ) √ u (5) Set a = X − f , b = X + f , and c = Y . Theorem 3.9.
Let f, γ, δ , and u be as in Algorithm 3.8. If ( a, b, c ) is generatedby Algorithm 3.8 then it is a PPT.Proof. That ( a, b, c ) is a PT is established in the paragraph preceding Theorem 3.2,along with Theorem 3.5 and Proposition 3.6.Suppose ( a, b, c ) is not primitive, and consider q ∈ Z with q | a , q | b , and q | c .As b − a = f and b + a = X , we have q = f and so f | X and f | Y . Now f β = γδ m u , for some β ∈ Z [ √ f = u ¯ u , it follows ¯ u β = γδ m u . As γ is aunit in Z [ √ u ∈ (¯ u ), i.e. p ⊂ ¯ p . But this is impossible for q = f ≡ ± (cid:3) Density of primitive ( a, b, b + g ) triples We aim to study the asymptotic density of the set of PPT of ( a, b, b + g ) formamong the set of all PPT with matching parity of a . In light of Theorem 1.2, theset of all PPT of ( r − s , rs, r + s ) form (and separately PPT with a = 2 rs ) isin bijection with the set of pairs ( r, s ) with s < r and gcd( r, s ) = 1. We consider P ( B ) = { ( r, s ) : gcd( r, s ) = 1 , < s < r, r ≤ B } ⊂ Z . We begin by recalling some classical number theoretic functions (see for example[1, § φ ( n ) is the number of k such that 1 ≤ k ≤ n andgcd( k, n ) = 1. Also, the M¨obius function µ ( n ) is defined by µ (1) = 1, µ ( n ) = ( − k for n = p · · · p k a product of distinct primes, and µ ( n ) = 0 if n has a square factor. Proposition 4.1. P ( B ) = π B + O ( B log B ) . Proof.
By definition of P ( B ), P ( B ) = X r ≤ B { ( r, s ) : gcd( r, s ) = 1 , < s < r } = X r ≤ B φ ( r ) . By [1, Theorem 3.7], P r ≤ B φ ( r ) = π B + O ( B log B ). (cid:3) In light of the proof of Theorem 2.4 and Corollary 2.5, we consider separatelythe cases g = m , with m odd; g = 2 m , with m even; and g = 2 m , with m odd. By the work in Proposition 2.2, these cases deal with PPT of the form( r − s , rs, rs + m ), (2 rs, r − s , r − s + 2 m ), and (2 rs, r − s , r − s + 2 m ),respectively. Thus, PPT in the first case will be compared with all PPT for which b = 2 rs , and the latter two cases will be compared with all PPT for which a = 2 rs . Example 4.2.
Let g = 1. By Corollary 2.5, a PPT ( a, b, b + 1) is obtained from r n = n + 1 and s n = n , for some n . We let G ( B ) = { ( n + 1 , n ) : n + 1 ≤ B } , andcompute lim B →∞ G ( B ) P ( B ) = lim B →∞ B π B = 0 . Our approach is different from [7, § a or b , and proves an asymptotic formula of Lehmer. ANDREW SCHMELZER AND SUNIL CHETTY
The example illustrates that PPT of the form ( a, b, b + g ), for a fixed g , may berare among all PPT, and so instead we consider the collection of all PPT of theform ( a, b, b + g ), for each family of values of g in Corollary 2.5.4.1. Case g = m odd. In this case, we know m is odd. By Corollary 2.5, thePPT for this case are given by r = ( k + m ) and s = ( k − m ) for some odd integer k , with gcd( k, m ) = 1. The set of such pairs ( k, m ) is in bijection with the set of( r, s ) described by the formulas and satisfying gcd( r, s ) = 1. Notice m < k ≤ B implies r ≤ B . Let G ( B ) = { ( k, m ) : gcd( k, m ) = 1 , < m < k ≤ B, k, m ≡ } ⊂ Z . Proposition 4.3.
For N ∈ Z + odd, there are φ ( N ) / odd integers ≤ m < N such that gcd( m, N ) = 1 .Proof. For each odd m with 1 ≤ m < N , with gcd( m, N ) = 1, we know N − m iseven, 1 < N − m ≤ N −
1, and gcd( N − m, N ) = 1. (cid:3) We define the parity function d by d ( n ) = 1 if 2 ∤ n and d ( n ) = 0 if 2 | n . Definition 4.4.
The 2-Euler totient function φ ( n ) is the product φ ( n ) = ( φ · d ) ( n )of the Euler totient function φ and the parity function d , so φ ( n ) = (cid:26) φ ( n ) if 2 ∤ n | n. Because φ and d are multiplicative functions, we know φ is a multiplicativefunction also. We set f ( n ) = P d | n φ ( d ) and apply the M¨obius inversion formula(see [1, Theorem 2.9]) to obtain φ ( n ) = X d | n µ ( d ) f (cid:16) nd (cid:17) . Lemma 4.5.
For n = 2 e u , with e ≥ and ∤ u , f ( n ) = u . In particular, if e = 0 then f ( n ) = n .Proof. All divisors d of 2 e u are of the form d = 2 a v with 0 ≤ a ≤ e and v | u .When e = 0, we also have a = 0 and so all factors of n are odd. Thus f ( n ) = X d | n φ ( d ) = X d | n φ ( d ) = n, with the right-most equality by [1, Theorem 2.2].For e >
0, we know φ ( d ) = 0 when 2 | d , and so P d | n φ ( n ) = P odd d | n φ ( d ).The odd divisors of n are the numbers v dividing u . Now f ( n ) = X d | n φ ( d ) = X odd d | n φ ( d ) = X v | u φ ( v ) = u, again using [1, Theorem 2.2] for the last equality. (cid:3) Theorem 4.6. P n ≤ B φ ( n ) = π B + O ( B log B ) . N PRIMITIVE PYTHAGOREAN TRIPLES OF SPECIAL FORMS 9
Proof.
The argument follows the same strategy as [1, Theorem 3.7]. By the M¨obiusinversion formula stated above, we have P n ≤ B φ ( n ) = P n ≤ B P d | n µ ( d ) f (cid:0) nd (cid:1) = P q,d, qd ≤ B µ ( d ) f ( q )= P d ≤ B µ ( d ) P q ≤ B/d f ( q )= P d ≤ B µ ( d ) (cid:16)P odd q ≤ B/d f ( q ) + P even q ≤ B/d f ( q ) (cid:17) . . By Lemma 4.5, and that the sum of the first N consecutive odds yields N , X odd q ≤ B/d f ( q ) = X odd q ≤ B/d q = (cid:18) B d (cid:19) + O (1) = B d + O (1) . For even q ≤ B/d we write q = 2 a j , with 2 ∤ j , and so using Lemma 4.5, P even q ≤ B/d f ( q ) = P M B a =1 P odd j ≤ B/ (2 a d ) f (2 a j )= P M B a =1 P odd j ≤ B/ (2 a d ) j = P M B a =1 (cid:0) B a +1 d (cid:1) + O (1)= B d (cid:16)P M B +1 a =0 (cid:0) (cid:1) a − (cid:17) + O (log B )= B d (cid:16) − (cid:0) (cid:1) M B +2 (cid:17) + O (log B ) , where M B = ⌊ log B ⌋ . Since (1 / log B +2 = 1 / (16 B ), we now have X even q ≤ B/d f ( q ) = B d + O (log B ) . With these two sums handled, we now have P n ≤ B φ ( n ) = P d ≤ B µ ( d ) (cid:16)P odd q ≤ B/d f ( q ) + P even q ≤ B/d f ( q ) (cid:17) . = P d ≤ B µ ( d ) (cid:16) B d + B d + O (log B ) (cid:17) = B P d ≤ B (cid:16) µ ( d ) d (cid:17) + O (log B ) P d ≤ B µ ( d )= B (cid:0) π + O ( B ) (cid:1) + O ( B log B ) , using [1, § µ ( d ). (cid:3) Theorem 4.7. lim B →∞ G ( B ) P ( B ) = . Proof.
From the definition of G ( B ) and Proposition 4.3, we have G ( B ) = { ( k, m ) : gcd( k, m ) = 1 , < m < k ≤ B, k, m ≡ } = P odd k ≤ B φ ( k )= P k ≤ B φ ( k ) . and thus from Theorem 4.6, G ( B ) = π B + O ( B log B ) . Combining this withProposition 4.1 gives the result. (cid:3)
Case g = 2 m even. In this case, we must consider both the case of m evenand of m odd. Recall, in the comments following Proposition 4.1, that in this casewe compare the PPT of ( a, b, b + 2 m ) form with all PPT of (2 rs, r − s , r + s )form.For m even, following with Corollary 2.5, we define G ( B ) = { ( k, m ) : gcd( k, m ) = 1 , < m < k ≤ B, m ≡ , k ≡ } ⊂ Z . Theorem 4.8. lim B →∞ G ( B ) P ( B ) = . Proof.
From the definition of G ( B ) and Proposition 4.3, we have G ( B ) = { ( k, m ) : gcd( k, m ) = 1 , < m < k ≤ B, m ≡ , k ≡ } = P odd k ≤ B φ ( k )= P k ≤ B φ ( k ) . and thus from Theorem 4.6, G ( B ) = π B + O ( B log B ) . Combining this withProposition 4.1 gives the result. (cid:3)
We now consider m odd. As above, we compare the PPT of ( a, b, b + 2 m ) formwith all PPT of (2 rs, r − s , r + s ) form. Following Corollary 2.5, we define G ( B ) = { ( k, m ) : gcd( k, m ) = 1 , < m < k ≤ B, m ≡ , k ≡ } ⊂ Z . Theorem 4.9. lim B →∞ G ( B ) P ( B ) = . Proof.
From the definition of G ( B ), we have G ( B ) = { ( k, m ) : gcd( k, m ) = 1 , < m < k ≤ B, m ≡ , k ≡ } = P even k ≤ B φ ( k ) , as each m with gcd( k, m ) = 1 is necessarily odd when k is even. Now, from X k ≤ B φ ( k ) = X even k ≤ B φ ( k ) + X odd k ≤ B φ ( k ) = X even k ≤ B φ ( k ) + X odd k ≤ B φ ( k ) , [1, Theorem 3.7] (as in Proposition 4.1), and Theorem 4.6, we obtain X even k ≤ B φ ( k ) = 3 π B − π B + O ( B log B ) = 1 π B + O ( B log B ) . (cid:3) Future Work
The case of ( a, a + f, c ) PPT has certainly not been treated as thoroughly as thecase of ( a, b, b + g ) PPT, and so the focus of future work will be those ( a, a + f, c )triples. In particular, in § a, a + f, c ) form and PPT of ( r − s , rs, r + s ) form (or flipped parity),as in Corollary 2.5.As a first step, if a = r − s then we have r − s + f = a + f = b = 2 rs. N PRIMITIVE PYTHAGOREAN TRIPLES OF SPECIAL FORMS 11
It follows that − f = r − rs − s = ( r − s ) − s , and so ( r − s, s ) are solutions to a similar Pell equation as considered in § a = 2 rs instead only changes the sign of f in the Pell equation. Thegoal in the end is to be able to establish density results for PPT of ( a, a + f, c ) formanalogous to § m (allowing for negative values also)and u . In addition, we have given no attention to the ± sign that comes along withsolutions to the classical negative Pell equation. It would be interesting to studyhow to construct ( a, a + f, c ) PPT with minimal a , and possibly to determine arelation between values f and corresponding minimal a . References [1] T.M. Apostol.
Introduction to Analytic Number Theory . Undergraduate Texts in Mathematics.Springer-Verlag, 1976.[2] H. Davenpot.
The Higher Arithmetic . Cambridge University Press, eighth edition, 2008.[3] J. Gallian.
Contemporary Abstract Algebra . Brooks/Cole, seventh edition, 2010.[4] K. Ireland and M. Rosen.
A Classical Introduction to Modern Number Theory , volume 84 of
Graduate Texts in Mathematics . Springer, 1990.[5] W.J. LeVeque.
Fundamentals of Number Theory . Dover Books on Mathematics. Dover, 1996.[6] The Sage Developers.
Sage Mathematics Software System (Version 8.7) , 2019. .[7] R. Takloo-Bighash.
A Pythagorean Introduction to Number Theory: Right Triangles, Sums ofSquares, and Arithmetic . Undergraduate Texts in Mathematics. Springer, 2018.
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