The proof,is divided in 4 sections...
x=2k+1 ,y=2h+1 with replacement we will finally have 4/m*((k-h)(k+h)+(k-h))=z ,and therefore k-h=sm and k+h=fm =>z=4s(fm+1) and finally x=m*(s+f)+1,y=m*(f-s)+1,z=4*s(fm+1)... –
we put..x=2k ,y=2h => z=(4/m)(k-h)(k+h)=4sfm, s,f,k,h in Z. Also we can to accept that k-h=sm and k+h=f,,and we have relations per case,i.e analytically i)z=4s(f+1),x=(sm+f)+1,y=(f-sm)+1 ii)x=sm+f and y=fsm, z=4sf. –
3.x=2k+1,y=2h and we have i) (2k+1)^2-(2h)^2=mz, i call 2k+1-2h=lm and finally z=l*(lm+4h) ii) 2k+1+2h=lm=> z=l(lm-4h) ,l,m,k,h in Ζ .. –
4.x=2k,y=2h+1 and we take i) (2k)^2-(2h+1)^2=mz,and also i call 2k-(2h+1)=lm and finally z=l*(2*(2h+1)+lm) ii) 2k+2h+1=lm => therefore z=l(lh-2(2m+1)) ,l,h,k,m in Ζ. – Nikos Mantzakouras
A solution with Mathematica Approach ... YouYou could try
FindInstance[{X^2 - Y^2 == M Z }, {X, Y, Z, M}, Integers, 3]
(*{{X -> 0, Y -> 301, Z -> -1, M -> 90601}, {X -> -139, Y -> -139,Z -> 1, M -> 0}, {X -> 8, Y -> 8, Z -> -37, M -> 0}}*)
The general solution can be calculated as follows:
Reduce[{x^2 - y^2 == m*z, Element[{x, y, m}, Integers] }, z, Integers]
(*(m | x | y | z) \[Element]Integers && ((m <= -1 && z == (x^2 - y^2)/m) || (m ==0 && (y == -Abs[x] || y == Abs[x])) || (m >= 1 &&z == (x^2 - y^2)/m))*)
