If $a \in \Bbb Z$ is the sum of two squares then $a$ can't be written in which of the following forms?

Question

Let $a \in \Bbb Z$ be such that $a = b^2 + c^2,$ where $b,c \in \Bbb Z \setminus \{0\}.$ Then $a$ cannot be written as $:$

$(1)$ $p d^2,$ where $d \in \Bbb Z$ and $p$ is a prime with $p \equiv 1\ \left (\text {mod}\ 4 \right ).$

$(2)$ $p q d^2,$ where $d \in \Bbb Z$ and $p,q$ are distinct primes with $p,q \equiv 3\ \left (\text {mod}\ 4 \right ).$

$(1)$ is false because $2^2 + 1^2 = 5 = 5 \cdot 1^2,$ where $d = 1 \in \Bbb Z$ and $5$ is a prime with $5 \equiv 1\ \left (\text {mod}\ 4 \right ).$ How do I prove or disprove the other option? Any help in this regard will be appreciated.

Thanks for your time.

Answer 1

Note the Sum of two squares theorem states

An integer greater than one can be written as a sum of two squares if and only if its prime decomposition contains no term $p^k$, where prime $p\equiv 3 \pmod{4}$ and $k$ is odd.

For your $(2)$, with $p$, since it's distinct from $q$, then the power of $p$ in $a$ would be $1$ plus $2$ times the power of $p$ in $d$, i.e., the exponent of $p$ is odd. Thus, since $p \equiv 3 \pmod{4}$, then the theorem quoted above says the value cannot be the sum of two squares.

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Note the Wikipedia's article's proof link to the Internet Archive says "This item is no longer available". I did some searching, but couldn't find any other link to it. However, there's basically the equivalent theorem stated in Sum of Two Squares near the bottom of page $4$:

Theorem $6$. A positive integer n is a sum of two squares iff $\operatorname{ord}_p(n)$ is even for all primes $p \equiv 3 \pmod{4}$.

It then follows with a remark about an equivalent statement (which is also equivalent to Wikipedia's theorem statement I quoted initially):

Remark: An equivalent statement of the theorem, which we will use in the proof, is: $n$ is a sum of two squares iff it factors as $n = ab^2$, where $a$ has no prime factor $p \equiv 3 \pmod{4}$.

This statement is possibly somewhat confusing as it's basically the equivalent of saying $a$ is square-free. Anyway, the linked paper goes on to state & prove a lemma it then uses to prove its theorem $6$.

Answer 2

There already is an accepted answer, but I wanted to point out a more self-contained argument. A way to prove 2) is to use the following elementary lemma: if $p=3$ mod $4$ is a prime and $a,b$ are integers such that $p\mid a^2+b^2$, then $p\mid a$ and $p\mid b$, thus $p^2\mid a^2+b^2$.

This lemma implies that for any prime $p=3$ mod $4$ and any integers $a,b$, $v_p(a^2+b^2)$ is even, which shows 2).

Now, how to prove the lemma?

Assume we have $p\mid a^2+b^2$ and, say, $p$ does not divide $a$. Let $a'$ be its inverse mod $p$; take $b'=ba'$. Then $p\mid b'^2+1$. As $\frac{p-1}{2}$ is odd, $(b')^2+1\mid (b')^{2\times (p-1)/2}+1$, so $p\mid (b')^p+b'$. But by Fermat's little theorem, $p|(b')^p-b'$ so $p\mid 2b'$. As $p \neq 2$, $p\mid b'$, so (as $a'$ is coprime to $p$) $p\mid b$. Therefore $p\mid a^2$ so $p\mid a$, a contradiction.