Pythagorean Triples : Is every positive integer $\gt$ $2$ part of at least one Pythagorean triple?

Using the characterisation of these triples, it suffices to show that any such number can be written as $m^2-n^2$, $2mn$ or $m^2+n^2$ with some numbers $m>n$.

The case $m^2-n^2$ covers "the most" numbers (only those $\equiv 2 \mod 4$ remain), the rest is covered by $2mn$.

I. Yes. Proof without words:

$$(\color{brown}{2m})^2+(m^2-1)^2 = (m^2+1)^2$$

$$(\color{brown}{2m+1})^2+(2m^2+2m)^2 = (2m^2+2m+1)^2$$

II. Higher.

To prove it for quadruples is easier since even and odd cases can be combined into a single identity,

$$n^2+(n+1)^2+(n^2+n)^2 = (n^2+n+1)^2$$

and for quintuples,

$$n^2 + (n-2)^2 + (2n+1)^2 + (3n^2+2)^2 = (3n^2+3)^2$$