Suppose that $p$ is a prime of the form $4k + 1$. I'm trying to prove that there exist nonzero integers $a$ and $b$ such that $p^2 = a^2 + b^2$. I begin by noting that $p$ itself can be written as the sum of two squares, since it does not have any prime factors of the form $4k+3$ (this is a theorem from my discrete math textbook: a positive integer $n$ can be represented as the sum of two squares if and only if every prime divisor of $n$ of the form $4k+3$ appears in the canonical representation of $n$ with an even exponent). We thus have $p=u^2+v^2$, and $p^2 = u^4+2u^2v^2+v^4$. I now need to prove that the right-hand side is the sum of two squares, but I'm lost. Obviously, it can be written as the sum of four squares, but that's not what I'm trying to show. Any hint or nudge in the right direction would be appreciated.
For any natural number $n\ge 1$, given pairs $(a_1,b_1),(a_2,b_2),...,(a_n,b_n)$ of integer numbers, there exist integer number $c$ and $d$ such that  $$\prod_{i=1}^{n}(a_i^2+b_i^2) = c^2+d^2$$    My initial approach is  Base Case: $(a_1^2+b_1^2) = a_1^2+b_1^2$ which is true. (Although it is trivial)  Prove the statement is true when $n=2$: We have $$(a^2+b^2)(c^2+d^2) = (ac-bd)^2+(ad+bc)^2$$ (Thanks André Nicolas for pointing it out)   So if $a,b,c,d$ are integers, $ac,bd,ad,bc$ are all integers and integers are closed under addition and subtraction. Hence $(ac-bd),(ad+bc)$ are integers.   Inductive Hypothesis: $\prod_{i=1}^{n}(a_i^2+b_i^2) = c^2+d^2$ is true  Inductive Step: $$\prod_{i=1}^{n+1}(a_i^2+b_i^2) = \prod_{i=1}^{n}(a_i^2+b_i^2)\cdot (a_{n+1}^2+b_{n+1}^2) = (c^2+d^2)\cdot (a_{n+1}^2+b_{n+1}^2)$$  Where $c$ and $d$ are integers.   But when we apply $n=2$, we have  $(c^2+d^2)\cdot (a_{n+1}^2+b_{n+1}^2) = (e^2 + f^2)$ where $e$ and $f$ are integers.  Hence, by the principle of induction, the statement we needed to prove is true.
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