## Equation: (x^2 + 1)(y^2 + 1) = (z^2 + 1)

Does there exist an infinite number of solutions to the equation:

$(x^2 \; + \; 1) \, (y^2 \; + \; 1) \; = \; (z^2 \; + \; 1)$

where each of   $x, \; y$   and   $z$   is a positive integer > 1

Here are two parametric solutions:

$x \; = \; n$
$y \; = \; n \; + \; 1$
$z \; = \; n^2 \; + \; n \; + \; 1$

$(n^2 + 1) \, ((n+1)^2 + 1) \; = \; (n^2 + n + 1)^2 \; + \; 1$

$x \; = \; n$
$y \; = \; 2 \, n^2$
$z \; = \; 2 \, n^3 \; + \; n$

$(n^2 + 1) \,((2 \, n^2)^2 + 1) \; = \; (2 \, n^3 + n)^2 \; + \; 1$

There may be other parametric solutions.

math grad - Interest: Number theory
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### 3 Responses to Equation: (x^2 + 1)(y^2 + 1) = (z^2 + 1)

1. paul says:

I’m going with yes.