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How can I convert this equation: $$321^2 - 196^2 = 64625$$ to be in this form: $$X^2 - Y^2 + X = 64625$$

Whereas $X$ and $Y$ are Odds and $X > \sqrt{64625}$

I tried to find $X$ value by testing the values from 255 until 321 and I found $X=257$ and $Y=41$. So I would like to asking the mathematicians if there is any mathematical solution to find $X$ and $Y$ values?

Thanks for your help.

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  • $\begingroup$ "Convert this equation" - what is that supposed to mean? The first equation has no relevance. $\endgroup$ – Peter Foreman Apr 16 at 20:56
  • $\begingroup$ Yes, it's equal but I just need to change the equation to be in the form $X^2 - Y^2 + X$ and I would like to know if there is any mathematical solution to do that ? Thanks $\endgroup$ – al3ndaleeb Apr 16 at 20:59
  • $\begingroup$ @PeterForeman Is't impossible ? or my question was wrong ? $\endgroup$ – al3ndaleeb Apr 16 at 21:10
  • $\begingroup$ I'm quite sure the solution you gave is the only one, but I don't know how to prove it. $\endgroup$ – Peter Foreman Apr 16 at 21:11
  • $\begingroup$ @PeterForeman Thanks alot for your reply. I will try to read and search and I hope some one guide me or correct my understanding. $\endgroup$ – al3ndaleeb Apr 16 at 21:15
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Times the equation by $4$, add 1 & complete the square \begin{eqnarray*} 4X^2+4X+1-4Y^2=4 \times 64625+1 \end{eqnarray*} which gives \begin{eqnarray*} (2X+1)^2-(2Y)^2=\color{red}{(2X-2Y+1)}\color{blue}{(2X+2Y+1)}=258501=\color{red}{433} \times \color{blue}{597}. \end{eqnarray*} etc ...

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    $\begingroup$ You can also use the factorization $$\color{red}{1}\cdot 258501.$$ That gives $X=Y= 64625.$ $\endgroup$ – Thomas Andrews Apr 16 at 21:50
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    $\begingroup$ ... or $1299\times199$ to get $(X,Y)=(374,-275)$, or $597\times433$ to get $(X,Y)=(257,-41)$ or $258501\times1$ to get $(X,Y)=(64625,-64625)$. $\endgroup$ – Servaes Apr 16 at 22:21
  • $\begingroup$ nice prove but this is in case X==Y. but what if X !=Y ?. I was thinking the differnce between two squares in the first equation will guide me to prove that. thanks for correction $\endgroup$ – al3ndaleeb Apr 17 at 9:38

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