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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=±12012-08-18
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    $a=\frac{b^2+1}{a^2+1}=\frac{(b^2+1)(a^2+1)}{(a^2+1)^2}=\frac{(ab±1)^2+(a∓b)^2}{(a^2+1)^2}=(\frac{ab±1}{a^2+1})^2+(\frac{a∓b}{a^2+1})^2$, so $a$ must be sum of two squares.2012-08-18
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    See http://math.stackexchange.com/questions/30501/integer-solutions-of-n2n1-m32012-08-18
  • 0
    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)$2012-08-18
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    It's just an elliptic curve, and there's a lot of theory about it. Try searching for integral points on elliptic curves.2012-08-18

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