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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.

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closed as off-topic by Michael E2, Mariusz Iwaniuk, b3m2a1, bbgodfrey, AccidentalFourierTransform Nov 3 '18 at 16:40

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "The question is out of scope for this site. The answer to this question requires either advice from Wolfram support or the services of a professional consultant." – Michael E2, Mariusz Iwaniuk, AccidentalFourierTransform
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 3
    $\begingroup$ Do you mean you want a purely mathematical solution rather than one using Mathematica? If that's the case try the Mathematics StackExchange where such problems are in scope. $\endgroup$ – b3m2a1 Nov 3 '18 at 15:30
  • $\begingroup$ Τo use the program is easy but does not solve the problem at its root. I could do it too. I will give it after a few hours just to be there.. $\endgroup$ – Nikos Mantzakouras Nov 3 '18 at 15:33
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    $\begingroup$ That sounds like this question should really be here. This site is solely intended for questions pertaining to Mathematica the software developed by Wolfram Research. Pure math questions are intended to be on math.stackexchange. $\endgroup$ – b3m2a1 Nov 3 '18 at 15:35
  • $\begingroup$ For reference, it would be quite useful to have a full example comprised of representative input (in Mathematica format so it can be cut-and-pasted), and the desired result. $\endgroup$ – Daniel Lichtblau Nov 4 '18 at 22:27
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You 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))*)
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  • 4
    $\begingroup$ @Nikos Mantzakouras Sorry, I rolled-back your edit because I falsely approved your edit. Now I am not convinced any more that your edit contained what the author intended. If you want to provide additional information, please edit your question. If you have found your own answer or an extension of this one, you may consider posting it as a separate answer. $\endgroup$ – Henrik Schumacher Nov 3 '18 at 13:41
  • $\begingroup$ sorry i dont want a approach solution i write in begin and i delete this because you delete my work $\endgroup$ – Nikos Mantzakouras Nov 3 '18 at 14:09
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    $\begingroup$ @NikosMantzakouras please take some time to read what SE is about: mathematica.stackexchange.com/help That way you won't be surprised that your edit to someone else's answer is not obligated to be approved. $\endgroup$ – Kuba Nov 3 '18 at 15:41
  • $\begingroup$ To write the work in mathematica more down (below) on mine .No on my work .I do not erase yours ..Better behavior more humane $\endgroup$ – Nikos Mantzakouras Nov 3 '18 at 15:44
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    $\begingroup$ @NikosMantzakouras To attempt to replace someone's answer by your own is considered to be very bad manners here. If you wish, submit your own answer using the box below. $\endgroup$ – bbgodfrey Nov 3 '18 at 16:28

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