Specific pythagorean triple A question states that $x,y,$ and $z$ are positive integers and $x^2 + 2021y^2 = z^2.$ It then proceeds to ask how many such triples $(x,y,z)$ there are. I would like to set $2021y^2 = a^2$ and than solve as a typical pythagorean triple, but I'm not sure how I should move on from there. Can someone give me a nudge in the right direction? Thanks.
 A: If you can find one solution, $(x,y,z)$, then the answer to the question is "infinitely many," because for any positive integer $a$, $(ax,ay,az)$ will be another.
So to find one solution note that $2021 = 43\cdot 47.$  The equation can be rewritten:
$$43\cdot 47y^2 = z^2-x^2 = (z-x)(z+x).$$
$43$ and $47$ divide the left side, so they must divide the right.  We guess that  that maybe $z-x=43$ and $z+x = 47.$  This gives $z=45$, $x=2$, forcing $y=1$.  So there's the one solution.  You can probably construct others similarly.
A: At least four distinct infinite families like Pythagorean triples. These are primitive, that is $\gcd(x,y,z) = 1,$ as long as $\gcd(u,v) = 1$ and we discard those additional cases, in the very first one we do not allow both $u,v$ odd for example.
$$ x = 47 u^2 - 43 v^2 \; , \; \; y = 2 u v \; , \; \;  z = 47 u^2 + 43 v^2  $$
$$ x = 2 u^2 + 90uv +2 v^2 \; , \; \; y = u^2 - v^2 \; , \; \;  z = 45 u^2 + 4uv + 45 v^2  $$
$$   x = 2 u^2 + 2uv -1010 v^2 \; , \; \; y = 2uv+v^2  \; , \; \;  z = 2 u^2 + 2uv + 1011 v^2   $$
$$   x =  u^2  -2021 v^2 \; , \; \; y = 2uv  \; , \; \;  z =  u^2 + 2021 v^2   $$
