Structure of primitive Pythagorean triples and the proof of a Fermat's theorem

Abstract

In a short survey of survey of primitive Pythagorean triples (x,y,z) 0 < x < y < z , we have found that one of x, y, z is divisible by 5 and z is not divisible by 3, there are Pythagorean triples whose corresponding element are equal , but there cannot be two Pythagorean triples such that (x10 y��" z1 ), (x1, zP z 2 ) , where z1 and z 2 hypotenuses of the corresponding Pythagorean triples. This is due to a Fermat's theorem [1] that the area of a Pythagorean triangle cannot be a perfect square of an integer, which can directly be used to prove Fermat's last theorem for n = 4. Therefore the preceding theorem is proved using elementary mathematics, which is the one of the main objectives of this contribution. All results in this contribution are summarized as a theorem. Theorem If (x, y, z) is a primitive Pythagorean triangle, where z is the hypotenuse, then z is never divisible by 3, andJ? = O(mod3) , xyz = O(mod5) ,and there are Pythagorean triangles whose corresponding one side is the same. But there are no two Pythagorean triangles such that (x1,y"z1) , (x"z"z 2 ) ), where z"z 2 are hypotenuses. Proof of the theorem Pythagoras' equation can be written as z2 =y2+x2 ,(x,y)= 1 (1) and if z = O(mod 3) ,then since J? is not divisible by 3, z2= y2 -1+ x2 -1+ 2 .Now, it follows at once from Fermat's little theorem that z cannot be divisible by 3. If xyz is not divisible by 5, squaring (1), one obtains z4 = y4 + x4 + 2x2 y2 and hence z4 -1 = y4 -1+ x 4 -1+ 2(x2 y2 ± 1)+ t, where t =- 1 or 3. Therefore xyz = O(mod5). It is easy to obtain two Pythagorean triples whose corresponding two elements are equal, from the pair-wise disjoint sets which have recently been obtained in Ref.2. For example 3652 =3642 +2 2 3652 =3642 +2 2 3652 =3572 +�� 2 .Now, assume that there exists ' ' two primitive Pythagorean triples of the form a z = bz + cz dz =az +cz (1) (2) It is clear that a is odd and c = O(mod 3) .F rom these two equations, one obtains immediately d2 - b2 = 2c2, d2 + b2 = 2a2, and therefore d4 -b4 = 4c2a2 = w�� (3) It has been proved by Fermat, after obtaining the representation of the primitive Pythagorean triples as x = 2rs,y = r2 -s2 ,z = r2 + s2, where 0 < s < r and r,s are of opposite parity, that (3) has no non trivial integral solution for d,b, w.To prove the same in an easy manner consider the equation d2 + b2 = 2a2 in the form d2 -a2 = a2 -b2 and writing it as ( d -a)( d + a) = (a -b)( a +b) use the technique used in Ref.3 to obtain the parametric solution for d andb If d -a= a -b., then d + b = 2a , from we deduce db= a2 .This never holds since (d,a)= 1 = ( b,a) by (1) and (2).1f (d -a)!!=( a -b) , where (u, v) =1 , V then V ( d + a)-= (a +b) . From these two relations, one derives the simultaneous u equations vd -ub = a( u -v) ud+vb=a(u+v) (4a) (4b) From ( 4a),( 4b ),it is easy to deduce the relations that we need to prove the theorem as (v2 + u2 )d = [2uv + u2 -v2 ]a, (v2 + u2 )b = [2uv -(u2 -v2 )]a, (v2 +u2)(d+b))=4uva, v2 +u2 =2a, assuming that uand v are odd. Hence d -b=(u2 -v2),(d+b)=2uv. Therefore d2 -b2 =2(u2 -v2)uv=2c2 and hence u, v are perfect squares and we can find two integers g,h such that. g4 -h4 = w21 < w�� .Now, proof of the last part of the theorem follows from the method of infinite descends of Fermat. Even if u and v are of opposite parity proof of the theorem can be done in the same way. To complete the proof of a Fermat's theorem that g4 -h4 = w�� is not satisfied by any non-trivial integers, we write (g2 + h2 )(g2 -h2) = w�� , where g,h are of opposite parity, to obtain g2 + h2 = x2, g2 -h2 = y2 and x4 -y4 = 4g2 h2 = z�� , where x and y are odd and eo-prime. But, in the case of the main theorem, we have shown that this is not satisfied by any non-trivial odd x, y and even z0 numbers . This completes the proof of the Fermat's theorem we mentioned above.