prove that $x^2 + y^2 = z^4$ has infinitely many solutions with $(x,y,z)=1$ 
Prove that $x^2 + y^2 = z^4$ has infinitely many solutions with $(x,y,z)=1$.

Do I use the terms $x= r^2 - s^2$, $y = 2rs$, and $z = r^2 + s^2$ to prove this problem?
Thanks for any help.
 A: Hint Let $w=z^2$. Then 
$$x^2+y^2=w^2$$
Solve this Pytagorean equation, and prove that infinitely many $w$ are perfect squares.
With your notation $w=r^2+s^2$ so you need $r,s$ to satisfy $r^2+s^2=z^2$...
A: Actually, a very strong statement is true, and quite easy: given any positive integer $n,$ such that all prime factors $p$ of $n$ satisfy $p \equiv 1 \pmod 4,$ then there is a solution to $$ x^2 + y^2 = n    $$ with $$ \gcd(x,y) = 1.$$ 
For $n$ a prime (so itself $1 \pmod 4$) the existence of a solution was finally proved by Euler. After that, there is a simple mathematical induction, using the two choices in someone's formula (Brahmagupta with $N=-1$), anyway
$$ (a^2 + b^2)(c^2 + d^2) = (ac+bd)^2 + (ad-bc)^2,   $$ or
$$ (a^2 + b^2)(c^2 + d^2) = (ac-bd)^2 + (ad+bc)^2  $$
with the induction hypothesis $$  \gcd(a,b) = \gcd(c,d) = 1.  $$ 
Indeed, given such an $n,$ with $r$ distinct prime factors and each $p_i \equiv 1 \pmod 4,$ but we otherwise ignore the exponents, the  number of representations $$  x^2 + y^2 = n  $$ with
$$  0 < x < y, \; \; \gcd(x,y) = 1   $$
is exactly $$  2^{r-1}. $$ For example, about ignoring exponents, we get
$$ 1 +4  = 5,  $$
$$ 9+16  = 25 = 5^2,  $$
$$ 4 + 121 = 125 = 5^3,  $$
$$ 49 + 576 = 625 = 5^4.  $$
Alright, nobody is paying attention to this one, but I thought I ought to make sure I knew how the induction proof went. It is easier if we demand the second factor prime, so we are taking
$$   a^2 + b^2 = n, \; \; \gcd(a,b) = 1,$$
$$  c^2 + d^2 = p, \; \; p \equiv 1 \pmod 4, \; \; \mbox{of course} \; \; \gcd(c,d) = 1.  $$ 
We are going to represent $np.$ One way is
$$ np = (ac+bd)^2 + (ad-bc)^2.   $$ Suppose these are not relatively prime, there is some positive prime $q$ such that
$$  q |  ac+bd \; \; \mbox{and} \; \;   q |  ad-bc.  $$
Multiply the first by $d$ and the second by $c$ and subtract, we get
$$ q | b (c^2 + d^2), \; \; \mbox{or} \; \; q | b p.   $$
Back to our pair, multiply the first by $c$ and the second by $d$ and add, we get
$$ q | a (c^2 + d^2), \; \; \mbox{or} \; \; q | a p.   $$
Since $ \gcd(a,b) = 1,  $ it follows that $q |p$ and then that $q=p.$
So the only possible GCD is a power of $p$ itself.  If, in addition, 
$$  p |  ac-bd \; \; \mbox{and} \; \;   p |  ad+bc,  $$
we get $ p | 2ac  $ and $p | 2bc,$ or $p | a$ and $p|b.$ This is false, as $\gcd(a,b) = 1.$
To put it briefly, the only possible obstacle to a proper representation of $np$ is a common factor of $p$ itself, but if $p$ divides all four of $ac+bd, \; ad-bc, \; ac-bd, \; ad+bc, $ then it divides both $a,b,$ a contradiction. So, at least  one of the four given expressions is not divisible by $p,$ and the pair using that expression gives a proper/primitive representation of $np$ as the sum of two squares. Naturally, the other expression in the successful pair is also prime to $p.$ 
