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Solve in positive integers $$a^2-b^2+4a=0$$

I tried considering the residues in mod4 but not so helpful. Any help/hint on how to approach this problem ? Thanks !

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HINT : $$\begin{align}\color{red}{a^2}-b^2\color{red}{+4a}=0&\iff \color{red}{(a+2)^2-4}-b^2=0\\&\iff (a+2)^2-b^2=4\\&\iff (a+2-b)(a+2+b)=4\end{align}$$

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  • $\begingroup$ thank you! should i be setting each factor on left hand side equal to the factors of 4 and solve a,b ? $\endgroup$ – drae Apr 27 '15 at 10:41
  • $\begingroup$ @drae: Yes, exactly. $\endgroup$ – mathlove Apr 27 '15 at 10:41
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    $\begingroup$ thanks a lot! looking exactly for a simple method like this $\endgroup$ – drae Apr 27 '15 at 10:42

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