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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=±1 –  lab bhattacharjee Aug 18 '12 at 16:22
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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. –  lab bhattacharjee Aug 18 '12 at 16:32
    
    
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
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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. –  Lieven Aug 18 '12 at 19:21

1 Answer 1

Using MAGMA and the method described by Pete L. Clark here one finds the complete set of integer solutions is

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

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The first and third entries are not solutions. –  Byron Schmuland Aug 18 '12 at 18:04
    
I posted the wrong list before. Here is the corrected list. –  jspecter Aug 19 '12 at 14:30
    
Well, I assumed it was a problem in using the software. Thanks for the update. –  Byron Schmuland Aug 19 '12 at 19:59

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