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 $n \in \mathbb{N}$?

For the special case when the number of is a square and is given us.
That is, in the equation:
$X^2+Y^2=Z^2+q^2$
where the number of $q$ - given us.
Then the solutions of the equation can be written ospolzovavshis Pell: $p^2-2k(k-1)s^2=\pm{q}$
$k$ - given us can be anything.
$X=p^2+2(k-1)ps+2k(k-1)s^2$
$Y=p^2+2kps+2k(k-1)s^2$
$Z=p^2+2(2k-1)ps+2k(k-1)s^2$
If we use the solutions of Pell's equation: $p^2-2s^2=\pm1$
Solutions have the form:
$X=sp^2+2qps\pm(qp^2+2(p+q)s^2)$
$Y=2qps+2s^3\pm(qp^2+2(p+q)s^2)$
$Z=sp^2+4qps+2s^2\pm(qp^2+2(p+q)s^2)$
