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

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 !

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}

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