aa r X i v : . [ m a t h . N T ] F e b A Theorem of Congruent Primes
Jorma Jormakka ,Sourangshu Ghosh
Abstract.
To determine whether a number is congruent or not is an oldand difficult topic and progress is slow.The paper presents a new theo-rem when a prime number is congruent number or not. The proof is notnecessarily any simpler or shorter than existing proofs, but the methodmay be useful in other contexts. The proof of Theorem 1 tracks the setof solutions and this set branches as a binary tree. Conditions set to thetheorem restricts the branches so that only one branch is left. Followingthis branch gives either a solution or a contradiction. In Theorem 1 it leadsto a contradiction. The interest is in the proof method, which maybe canbe generalized to non-primes.
Key words:
Congruent numbers, elliptic curves, number theory.
A positive integer which can be written as the area of a right triangle with threerational number sides is called a Congruent Number[1]. Alternatively it can bedefined as the numbers ( a, x, y, z, t ) such that the following condition hold: x + ay = z x − ay = t From this tuple ( x, y, z, t ), we can also derive the sides of the right angle triangle a, b, c such that a + b = c , and ab/ n by substituting a = ( y − z ) /t, b = ( y − z ) /t, c = 2 x/t A problem of significant interest is to determine whether a given natural number n can be the area of a right-angled triangle with rational number sides.This problemcan be alternatively said of the existence of rational points on some elliptic curvesthat are defined over Q .Note that if we multiply each side of the triangle whose area is a congruentnumber q by s , then it is evident that s q is also a congruent number for any nat-ural number s . Therefore a residue of the number q in the group Q ∗ /Q ∗ decideswhether the number q will be a congruent number or not. For this reason we onlyconsider square-free positive congruent numbers. An easy to way to determinewhether a given rational number is a congruent number is the Tunnell’s theoremnamed after number theorist Jerrold B. Tunnell who demonstrated the methodin [2].For a given square-free integer n , define the following numbers. A n = (( x, y, z ) ∈ Z | n = 2 x + y + 32 z ) ,B n = (( x, y, z ) ∈ Z | n = 2 x + y + 8 z ) ,C n = (( x, y, z ) ∈ Z | n = 8 x + 2 y + 64 z ) ,D n = (( x, y, z ) ∈ Z | n = 8 x + 2 y + 16 z ) , Now if n is actually a congruent number then by Tunnell’s theorem we haveif n is odd then 2 A n = B n and if n is even then 2 C n = D n . There have manyworks to classify congruent numbers especially for primes. Gross[3] proved thatif n is square free integer and has at most two prime factors of the form 5,6 or7 (mod 8), then n must be a congruent number. Monsky[4] proved the followingimportant properties to determine whether a number is congruent.The followingare all congruent numbers:1. p , p , p , also proved by Stephens[5]2. 2 p also proved by Heegner[6], and Birch (1968)[7,8] p p , p p , p p , p p p p provided ( p /p ) = − p p provided ( p /p ) = − p p , p p provided ( p /p ) = − p k refers to an arbitrary prime congruent to k mod 8 and ( a/p ) (wheregcd( a , p )=1) is the Legendre symbol which is 1 and -1 if a is a quadratic residueof p or not respectively. Iskra[9] proved the following important properties todetermine whether a number is not congruent. The following are not congruentnumbers:1. p , p , p q , p q also proved by Genocchi[10].2. p p ..p t provided ( p m /p n ) = − m < n .Lagrange[11] similarly proved the following important properties for a number tobe a non-congruent number1. p p provided ( p /p ) = − p p provided ( p /p ) = − n = p p q with the condition that n can be written as n = pqr or 2 pqr suchthat ( p/q ) = ( p/r ) = − n = 2 p p q with the condition that n can be written as n = pqr or 2 pqr suchthat ( p/q ) = ( p/r ) = − p ≡ p is a congruent number.2. n = p , p = a + 4 b , (( a + 2 b ) /p ) = − k in the congruence classes 5, 6, 7(mod 8),there are infinitely many square-free congruent numbers with k primefactors. Consider an elliptic curve of the form: y = x − d x (1)where d is an integer. A rational solution ( x, y ) to the elliptic curve (1) is a solutionwhere x and y are rational numbers.The substitution x = d ( a + b ) /b , y = 2 d ( a + c ) /b changes y = x − d x to a + b = c with ab = 2 d . Then 4 d = a ( c − a ). Integers d that give rationalnumber solutions to a + b = c , ab = 2 d are called congruent numbers. If d isa congruent number the elliptic curve (1) has a rational solution where y is notzero. In that case it has infinitely many rational solutions.If there is a solution for d = s , then there is a solution for d = 1 becausethe substitution y = s y ′ , x = s x ′ changes y = x − d x to y ′ = x ′ − x ′ . Itis known that no d = s is a congruent number as proved by Fermat using hismethod of infinte descent. The case where d is a prime number is almost solved.For notations the following concepts suffice: The condition that the integer a divides integer b is written as a | b . If p > p is denoted by Z p and Z ∗ p = { , . . . , p − } . The set of quadratic residuesmodulo p is the set QR p = { x ∈ Z ∗ p |∃ y ∈ Z ∗ p such that y ≡ x (mod p ) } . The set of quadratic nonresidues modulo p is the set QN R p = { x ∈ Z ∗ p | x QR p } . Let us start by two very simple lemmas.
Lemma 1.
Let c = a + b , a, b, c ∈ Z , then ∃ h, m, e ∈ IN such that a = ± hem , b = ± h ( m − e ) , c = ± h ( m + e ) . Proof.
Without loss of generality we can assume that a, b, c ∈ N . We can write c − b = ( c − b )( c + b ) = a . Let h = gcd ( c + b, c − b ). Then there exists m and e , m > e , gcd ( m, e ) = 1, such that c + b = hm , c − b = he . The claim follows.With Lemma 1 we can characterize congruent numbers. Lemma 2.
Let d ∈ Z , d > . Rational solutions ( x, y ) with x = 0 , y = 0 to y = x − d x are of the form ( x , y ) = (cid:18) d m + em − e , ± kj d m + em − e (cid:19) , ( x , y ) = (cid:18) d m − em + e , ± kj d m − em + e (cid:19) , where k, j, e, m ∈ N , m > e , gcd ( m, e ) = 1 , gcd ( k, j ) = 1 , satisfy d = (cid:18) k j (cid:19) m − e em . (2) Proof.
Let x, y ∈ Q , x = 0 , y = 0. Let us write α = dx + 1 ∈ Q , β = yx ∈ Q .Solving (10) for x and solving x from the definition of α yields x = β α − α = dα − . Writing β = kj for some k, j ∈ N gives α , = 1 − k j d ± p (2 dj ) + ( k ) j d . As y = 0, k = 0. By Lemma 1, α , ∈ Q if and only if there exist h, e, m ∈ N , gcd ( e, m ) = 1, m > e , such that k = hem , dj = 12 h ( m − e ) , c = 12 h ( m + e ) . If em = 0, then k = 0 and y = 0. This solution gives j = 2 dj α , = 1 ± dj dj = 1 ± , α = 2 , α = 0 ,x = dα − d , x = − d , y = 0but we have excluded this case in the assumptions. Since em = 0, let us write h = k em . Eliminating h yields d = (cid:18) k j (cid:19) m − e em ,c = k m + e ) . Simplifying α , yields α , = 1 m − e (cid:0) m − e − em ± ( m + e ) (cid:1) , i.e., α = 2 mm + e , α = − em − ex = dα − d m + em − e , x = − d m − em + e ,y = βx , β = (cid:18) kj (cid:19) = 4 d emm − e . This gives the claim.
As two examples of Lemma 2 d = 5 = (cid:18) · (cid:19) − · d = 7 = (cid:18) · (cid:19) − · gcd ( k, j ) = 1 but it is allowed that 2 | k .If d is a square, there are no rational solutions to (1) with y = 0. There arethe three solutions (0 , , ( d, , ( − d,
0) to (1), so the number of rational solutionsof (1) is finite, the rank of the elliptic curve is zero.In the next theorem gives a set of values where d is a prime number and (1)has no rational solutions, i.e., the elliptic curve has rank zero. The case of primenumbers d is rather well known: if p ≡ d ) or p ≡ d ) the number d is a congruent number and there are solutions to (1). If p ≡ d ) thereare no solutions and d is not a congruent number. The only case remaining is p ≡ d ). For that case it is known that e.g. p = 41 is a congruent number,while e.g. p = 17 is not.The next theorem does not solve the problem for any prime p that is onemodulo eight because if p ≡ − ∈ QR p , i.e., − ∈ QR p is equivalent with the condition that 4 | ( p −
1) and if p ≡ d ),then 8 | ( p − p = 19 is not a congruentnumber, but as 19 ≡ d ) this is known. Yet, the method of this proofseemed interesting enough to me in order to be written down. The method maygeneralize to other numbers than primes. The primality condition is used only ina few places. The main idea is to exclude branches from a recursion. Theorem 1.
Let d > be a prime such that − ∈ QN R d and ∈ QRN d .The equation (2) in Lemma 2 does not have solutions k, j, m, e ∈ IN where gcd ( m, e ) = 1 , gcd ( k, j ) = 1 , m > e > . Proof.
We write (2) with m , e d = (cid:18) k j (cid:19) m − e e m (3)If d | m then d | e and gcd ( m , e ) = 1, thus d m and d e . If d | k then since d is a prime d | k . It follows that k = dk and as gcd ( k, j ) = 1 holds d j . Thus(2 j ) m e = dk ( m − e )which is not possible as the left side is not divisible by d . Thus d k . Therefore d | m − e .If 2 k we convert (3) into the form d = (cid:18) kj (cid:19) stm − e (4)by the substitution m = m + e , e = m − e , i.e., 2 m = m + e , 2 e = m − e . As m e = ( m + e )( m − e ) = m − e holds em = ( m + e )( m − e ). As 4 | ( m − e )in (3) if 2 k it follows that one of m + e or m − e is even. If so, they areboth even and 2 | m + e , 2 | m − e and m, e are integers. As gcd ( m , e ) = 1, gcd ( m + e , m − e ) = 2. Then gcd ( m, e ) = gcd ((( m + e ) / m − e ) / m > e > m > e > | k then the substitution is m = m + e , e = m − e . Then m, e are integersand m > e >
0. In this case 2 j gecause gcd ( k, j ) = 1. Therefore 2 ( m − e ).It follows that gdc ( m, e ) = gcd ( m + e , m − e ) = 1. We get the same form (4)since me = m − e and m − e = 4 m e .Then d | em and j | em . Let us write (4) as j ( m + e )( m − e ) d = k me. (5) Since gcd ( m, e ) = 1 it follows that gcd ( m ± e, m ) = 1. Indeed, if m ± e = c r , m = c r for some r, c , c ∈ IN, then c c r = c m ± c e = c m ⇒ ( c − c ) m = ± c e ⇒ m | c ⇒ ∃ α ∈ IN such that c = αm ⇒ m = αmr ⇒ αr = 1 ⇒ r = 1 . Similarly, gcd ( m ± e, e ) = 1.Since gcd ( k, j ) = 1 it follows from (4) that k = m − e . Therefore (4) impliesthat dj = em . As dj = em and gcd ( e, m ) = 1 there is one of the cases: either m = ds , e = t for some s, t > m = s , e = dt .As k = ( m + e )( m − e ) and gcd (( m + e )( m − e )) ≤ m + e = c and m − e = c for some c , c > m + e = 2 c and m − e = 2 c .We have four cases in total.Case 1. m = ds , e = t , m + e = c , m − e = c . Then m − e = s d − t = c . The equation yields − ≡ ( c t − ) (mod d ) which is impossible since − ∈ QN R d .Case 2. m = ds , e = t , m + e = 2 c , m − e = 2 c . Then s d + t = 2 c , s d − t = 2 c . Multiplying the modular equations t ≡ c (mod d ) , − t ≡ c (mod d ) yields − ≡ (2 c c t − ) (mod d ) which is impossible since − ∈ QN R d .Case 3. m = s , e = dt , m + e = c , m − e = c . Then s + t d = c , s − t d = c . Thus 2 s = c + c (6)so 4 s = c + 2 c c + c + c − c c + c (2 s ) = ( c + c ) + ( c − c ) . (7)It follows from Lemma 1 that ∃ h ′ , e ′ , m ′ ∈ IN, gcd ( m ′ , e ′ ) = 1 such that c + c = h ′ e ′ m ′ , c − c = 12 h ′ ( m ′ − e ′ ) , s = 12 h ′ ( m ′ + e ′ ) . Solving c , c , s yields c = 14 h ′ (2 e ′ m ′ + m ′ − e ′ ) ,c = 14 h ′ (2 e ′ m ′ + e ′ − m ′ ) ,s = 14 h ′ ( m ′ + e ′ ) . Since 2 t d = c − c = ( c − c )( c + c )we get d = 14 t h ′ e ′ m ′ ( m ′ − e ′ ) i.e. d = (cid:18) h ′ e ′ m ′ t (cid:19) ( m ′ − e ′ ) e ′ m ′ . Removing the greatest common divisor of h ′ e ′ m ′ and t this equation can be writtenas d = (cid:18) k i +1 j i +1 (cid:19) ( m i +1 − e i +1 ) e i +1 m i +1 . (8)As gcd ( m ′ , e ′ ) = 1 and we made gcd ( k, j ) = 1, equation (8) is is of the same formas (3) d = (cid:18) k i j i (cid:19) ( m i − e i ) e i m i = (cid:18) k j (cid:19) ( m − e ) e m . We have a recursion that in each step reduces the numbers m i , e i to numbers m i +1 , e i +1 that are of the order of square root of m i , e i .Case 4. m = s , e = dt , m + e = 2 c , m − e = 2 c . We can select c > c ≥ s + t d = 2 c , s − t d = 2 c . Thus s = c + c dt = c − c = ( c − c )( c + c ) . (9)Let us notice that m + e = 2 c and1 = gcd ( m + e, e ) = gcd (2 c , dt ) ⇒ gcd ( c , t ) = 1 , gcd (2 , t ) = 11 = gcd ( m − e, e ) = gcd (2 c , dt ) ⇒ gcd ( c , t ) = 1 . First we exclude one case in the second equation of (9). If t > c + c = α t and c − c = α t for some α , α ∈ IN, then2 c = ( α + α ) t ⇒ t = 1 , c = α + α , c = ( α − α ) t ⇒ t = 1 , c = α − α . Thus, dt = c − c = α α t . It follows that d = α α and as d is prime andnecessarily α > α it follows that α = d , α = 1. Then c = d + 1 and c = d − s = c + c = 2( d −
1) is even, so m is even. Since s + dt = 2 c it would follow that t is also even as d is odd, but t = 1 in this case. We have acontradiction.Thus, in (9) must be one of the three cases t | ( c + c ) ⇒ ( c − c ) | d ⇒ c − c = d ⇒ t = c + c , or t | ( c − c ) ⇒ ( c + c ) | d ⇒ c + c = d ⇒ t = c − c , or t = 1 . In the first case 2 c = t + d ≥ , c = t − d ≥ . In the second case 2 c = d + t ≥ , c = d − t ≥ . In both of these two cases we can derive in a similar way: s = c + c ⇒ (2 s ) = (2 c ) + (2 c ) yields (2 s ) = ( d + t ) + ( d − t ) . (10)By Lemma 2 there exist h ′ , e ′ , m ′ ∈ IN such that d + t = h ′ e ′ m ′ , d − t = 12 h ′ ( m ′ − e ′ ) . The first equation implies that d h ′ . Thus4 d = h ′ (( m ′ + e ′ ) − e ′ )i.e., as h ′ d ) 2 ≡ ( m ′ + e ′ ) e ′− (mod d ) (11)which is a contradiction since 2 ∈ QN R d . There remains the case t = 1. Then2 c = s + d , 2 c = s − d . Instead of (10) we get(2 s ) = ( d + s ) + ( d − s ) . The contradiction (11) comes in the same way with t replaced by s . This meansthat Case 4 is not possible.Because Cases 1, 2 and 4 are not possible, only Case 3 is left. Case 3 gives arecursion formula. The values h ′ , m ′ , e ′ in Lemma 1 satisfy e ′ m ′ = ab + c = c − bah ′ = gcd ( b + c, b − c )giving a = c − b . The numbers h ′ , m ′ , e ′ can be chosen to be positive and onthe order of a, b, c . Thus, h ′ , m ′ , e ′ in (8) are of the order c , c . The numbers c , c are of the order √ m , √ e . Therefore in each step the numbers m i , e i get smaller,they are reduced to the order of their square roots. Consider the problem whenthe recursion stops.Let us look at an example of d = 5. Then d = 5 = (cid:18) · (cid:19) − · . We have m = 9 , e = 1 , k = 3 , j = 2. We can do the first step and find m =5 , e = 4 and d = 5 = (cid:18) (cid:19) · − . Identifying k = 3 = 5 − = 9, j d = 4 · · me , m = ds = 5 · , e = t = 2 , m + e = 5 + 4 = 3 = c and m − e = 5 − = c shows thatthe logic in the lemma is correct. We have Case 1, but for d = 5 the conditionsof the lemma are not fulfilled: − ∈ QN R . This is why Case 1 does not give acontradiction. What happens in Case 1 is that when we remove the term dt in acase resembling (6) we do not get (6) but2 t = c − c Therefore we do not get (7) which can be inserted to the equation to Lemma 1for calculation of the numbers h ′ , m ′ , e ′ .Let us look at another example, that of d = 7. Here − ∈ QN R and the Caseis not 1. d = 7 = (cid:18) · (cid:19) − · . We have m = 16 , e = 9 , k = 24 , j = 5. We find m = 16 + 9 = 25 , e = 16 − d = 7 = (cid:18) (cid:19) · − . Here k = 24 = 576 = 25 − = m − e , j d = 25 · · me , m = s = 5 , e = dt = 7 · , m + e = 25 + 7 = 32 = 2 · = 2 c and m − e = 25 − · = 2 c . The Case is 4. We notice that t = 1 and c = 4, c = 3, thus we have the case t = 1. Then s + d = 5 +7 = 32 = 2 · = 2 c and s − d = 5 − · = 2 c . We get(2 s ) = 100 = 64 + 36 = (2 c ) + (2 c ) = (5 + 7) + (5 − and therefore find the numbers h ′ , m ′ , e ′ for 10 = 8 + 6 . The numbers are h ′ = gcd (10 + 6 , −
6) = 4, e ′ = 1, m ′ = 2. Thus d + t = h ′ e ′ m ′ = 7 + 1 = 8 , d − t = 12 h ′ ( m ′ − e ′ ) = 6are true and 4 d = h ′ (( m ′ + e ′ ) − e ′ ) = 28 = 4 · (3 − . We get the modular equation 3 ≡ ∈ QN R d , but indeed 2 ∈ QR . Therefore for d = 7 we do not get a contradiction.The way the lemma works is that in (2) the numbers m and e must besquares m = s , e = t so that k can cancel them. The condition − ∈ QN R d excludes the larger branch ( s + t ) of m − e = ( s + t )( s − t )by ( s + t ) ≡ d ) being impossible.Therefore 4 d | ( m − e ) leads to 4 d | ( s − t ). The condition 2 ∈ QN R d excludesCase 4 and leaves only Case 3 which gives a recursion. Thus, the numbers m i , e i get smaller.If there is a congruent number d with − ∈ QN R d , the recursion must continueuntil it stops in some way and not to a contradiction, but the recursion does notstop and continues to a contradiction. At each stage 4 d | ( m i − e i ) or d | ( m i − e i )depending on if k i is odd or even. The numbers m i and e i become smaller on eachstep. Finally we must have 4 d = m i − e i or d = m i − e i .Changing variables in (2) to m = ( m i + e i ) / e = ( m i − e i ) / k is odd and m = m i + e i , e = m i − e i if k is even we get d = k j mem − e . (12) When the recursion has reached 4 d = m i − e i or d = m i − e i the number j = 1.In (12) necessarily k = m i e i and consequently d = me . As d is prime either m = d , e = 1 or m = 1, e = d . As in Cases 1 and 2 the choice m = d leads to − ∈ QR d and is impossible. Thus m = 1 and t = d , but then m − e < d > d filling the conditions of the lemma: for d = 19 holds − ∈ QN R and 2 ∈ QN R . We also get a small result: Corollary 1. If p is a prime and p ≡ , then ∈ QR p . Proof. If p is a prime and p ≡ p is a congruent number. Thereforethe conditions of Theorem 1 cannot be fulfilled. The condition − ∈ QR p isequivalent with 4 | ( p − p − k for some k , it follows that 4 ( p − − ∈ QN R p . The only other condition in Theorem 1 is that 2 ∈ QN R p .This can be otherwise be proved easily without using the Theorem 1 by Gauss’lemma which states that ( a/p ) = ( − n . Here ( a/p ) (where gcd( a , p )=1) is theLegendre symbol which is 1 and -1 if a is a quadratic residue of p or not respec-tively. Here n is the number of integers in the set S = ( a, a, a, ...., (( p − / a )whose remainder will be greater than p/ p . Putting the valueof a as 2 we get (2 /p ) = ( − n , where n is the number of integers in the set S = (1 , . , . , ...., (( p − / . S is smaller than p . Therefore theproblem reduces to only count the number of elements that exceed p/ n = ( p − / − [ p/ p is a prime and p ≡ we have p of the form 8 k + 7. Therefore n = (8 k + 7 − / − [8 k + 7 /
4] = 4 k + 3 − (2 k + 1) = 2 k + 2As n is even, we have 2 ∈ QR p . Whether primes are congruent numbers of not is an old and difficult topic andprogress is slow. The paper presents a new proof to a known theorem. The proofis not necessarily any simpler or shorter than existing proofs, but the method maybe useful in other contexts. The proof of Theorem 1 tracks the set of solutionsand this set branches as a binary tree. Conditions set to the theorem restrictsthe branches so that only one branch is left. Following this branch gives eithera solution or a contradiction. In Theorem 1 it leads to a contradiction. Usingdifferent conditions in this method may give new results.
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