How to solve $x^2+y^2=pz^2$ in $x,y,z\in\mathbb{N}$ if $p$ is such a prime that $p\equiv 3 \pmod 4$?
For such a prime we have lemma: $$p\mid a^2+b^2 \implies p\mid a\;\; {\rm and}\;\; p\mid b$$
Let's go to the problem:
Since $p\mid x^2+y^2$ we get $p\mid x$ and $p\mid y$ so $x=px'$ and $y=py'$ and now we have $$p^2x'^2+p^2y'^2 = pz^2\implies p\mid z $$ so $z=pz'$ and we get $$x'^2+y'^2=pz'^2$$ but this is the same equation as before with $x'$ smaller then $x$ and so on. So we can repeat this infinite times but this is impossible. So $x=y=z=0$.
Well, I am not sure either way about your lemma. The thing is though, there is a fairly short proof of the statement you are trying to show, that does not assume the lemma [which, if this were a homework question, would be what I think the instructor would want you to do].
The proof I had in mind:
If $x^2+y^2 = pz^2$, then we can assume that not all of $x,y,z$ are even [make sure you see why]. Consider two cases:
Case 1: $x,y$ odd. Then $x^2+y^2 \equiv 2$ mod 4, which implies that $pz^2 \equiv 2$ mod 4 which implies $z^2 \equiv 2$ mod 4 as $p \equiv 3$ mod 4. This is impossible as the only squares mod 4 are 0 or 1.
Case 2: $x$ odd, $y$ even. Then $x^2+y^2 \equiv 1$ mod 4 [why?] which implies $pz^2 \equiv 1$ mod 4, but as $p \equiv 3$ mod 4 implies $z^2 \equiv 3$ mod 4. This cannot be either, as the only squares mod 4 are 0 or 1.
Case 3: $x,y$ even. Then $pz^2$ must be even implying $z^2$ must be even, which contradicts not all of $x,y,z$ even.