# Diophantine equation $a^3+a=b^2+1$

I have this Diophantine: $$a^3+a=b^2+1$$ I found $a=2$, $b=3$ works. Also $a=13$ , $b=47$ works. How can I find all the integer solutions?

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Clearly, b is odd. Another solution a=1,b=±1 –  lab bhattacharjee Aug 18 '12 at 16:22
$a=\frac{b^2+1}{a^2+1}=\frac{(b^2+1)(a^2+1)}{(a^2+1)^2}=\frac{(ab±1)^2+(a∓b)^2}{‌​(a^2+1)^2}=(\frac{ab±1}{a^2+1})^2+(\frac{a∓b}{a^2+1})^2$, so $a$ must be sum of two squares. –  lab bhattacharjee Aug 18 '12 at 16:32
–  jspecter Aug 18 '12 at 16:47
As b is odd, $b^2≡1(mod\ 8)=>b^2+1≡2(mod\ 8)$. Now $a^3+a(modulo\ 8)$ are 2 only for $a≡1,2 or 5(mod\ 8)$ –  lab bhattacharjee Aug 18 '12 at 16:49
It's just an elliptic curve, and there's a lot of theory about it. Try searching for integral points on elliptic curves. –  Lieven Aug 18 '12 at 19:21

$(1 , -1), (2 , 3), (13, -47), (1 , 1), (2 , -3), (13 , 47)$