Quadratic Diophantine Equations I note that the Diophantine equation, $x^2 + y^2 = z^2$, with $x, y, z \in \mathbb{N}$, has infinitely many solutions. Indeed, $(x, y, z) = (3,4,5)$ provides a solution, and for any $k \in \mathbb{N}$ : $(kx, ky, kz ) = (3k, 4k, 5k)$ provides a solution.
However, assuming $x, y, z \in \mathbb{N}$ with $x, y > 1$, is the same true for the Diophantine equations, 
$x^2 +y^2 = z^2 + 1$, 
$x^2 + y^2 = z^2 + 2$,
$x^2 + y^2 = z^2 + 3$
and more generally, for $x^2 + y^2 = z^2 + n$, for any $n \in \mathbb{N}$?
In particular, are there infinitely-many triples $(x, y, z) \in \mathbb{N}^3$ for which  $x^2 + y^2 = z^2  + n$ is true for infinitely-many values of $n \in \mathbb{N}$?

$x^2+y^2 = z^2+n$ is equivalent to $x^2-n = (z-y)(z+y)$.
Any composite odd number can be written as $(z-y)(z+y)$ for some integers $z$ and $y$, so it is enough to show that $x^2-n$ contains infinitely many composite odd numbers.
If $n$ is odd, then you can simply pick $x = 2kn$ for any $k$. Then $x^2-n$ is a multiple of $n$ and odd, which gives you two integers $y$ and $z$ satisfying the equation. You end up with the family of solutions $(2kn, 2k^2n-(n+1)/2, 2k^2n-(n-1)/2)$
If $n$ is even, then you can simply pick $x=2k(n-1)+1$ for any $k$. Then $x^2-n \equiv 1^2-1 = 0 \pmod {n-1}$, and it is odd so again this gives you two integers $y$ and $z$ satisfying the equation. You end up with the family of solutions $(2k(n-1)+1,2k^2(n-1)+2k-n/2,2k^2(n-1)+2k-1+n/2)$
