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How do you find all positive integers $a,b,$ and $c$ such that $(a^2+1)(b^2+1)=c^2+1$?

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    b=a+1, c=ab+1 gives an infinite class of solutions. This can be found from (a+i)(a-i)(b+i)(b-i)=(c+i)(c-i) by choosing (a-i)(b+i)=c+i. I do not know whether there are other solutions. http://www.wolframalpha.com/input/?i=%28a^2%2B1%29*%28a^2%2B2*a%2B2%29-%28a^2%2Ba%2B1%29^22011-04-09
  • 1
    This problem is really bugging me, if $(c+i)(c-i) = x \bar{x}$ we can't say that $x = c\pm i$ so it seems like Gauss integers are basically useless for this problem.2011-04-09
  • 0
    Dear Amir, please consider adding some explanation or motivation to your questions.2011-04-09
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    This is also equivalent to $a^2b^2+a^2+b^2=c^2$. I.e., (ab,a,b,c) is a Pythagorean quadruple. Several parametrizations of Pythagorean quadruples (x,y,z,w) are known, but I do not think that adding the condition x=yz will bring something reasonable. http://en.wikipedia.org/wiki/Pythagorean_quadruple2011-04-10
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    @Martin, a couple of smallish solutions which aren't $b=a+1$ are $1,12,17$ and $2,8,18$.2011-07-18

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