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Solving Diophantine equations 5

Given a positive integers x,y,m would like to be able to find integer solutions z from Diophantine equation x^2-y^2 = m*z in Z.

---The proof,is divided in 4 sections---...

  1. x=2*k+1 ,y=2*h+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=4*s(fm+1) and finally x=m*(s+f)+1,y=m*(f-s)+1,z=4*s(fm+1)... –

  2. we put..x=2*k ,y=2*h => z=(4/m)(k-h)(k+h)=4*sfm, 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=4*s(f+1),x=(sm+f)+1,y=(f-sm)+1 ii)x=sm+f and y=fsm, z=4*s*f.

3.x=2*k+1,y=2*h and we have i) (2*k+1)^2-(2*h)^2=mz, i call 2*k+1-2*h=lm and finally z=l*(lm+4*h) ii) 2*k+1+2*h=lm=> z=l*(l*m-4*h) ,l,m,k,h in Ζ .

4.x=2*k,y=2*h+1 and we take i) (2*k)^2-(2*h+1)^2=mz,and also i call 2*k-(2*h+1)=lm and finally z=l*(2*(2*h+1)+lm) ii) 2*k+2*h+1=lm => therefore z=l*(l*h-2(2*m+1)) ,l,h,k,m in Ζ.