Product of $(4k-1)$ primes can't be sum of 2 squares

Question

I am trying to prove, Product of primes of the form $(4k-1)$ can't be sum of 2 squares. My approach is-

Let the product is $M=m_1m_2...m_n$ where $ m_1, m_2, ...m_n$ are primes.

Assume, $M$ can be written as sum of 2 squares. Then,

$M= x^2 + y ^2 \implies m_1m_2...m_n = (x+iy)(x-iy)$

But I am stuck at this stage, there must be something related to the property of Gaussian Integer.

How can it be proved?

Answer 1

Hint:

If there is a prime $p\equiv_4 3$ and $p\mid x^2+y^2$ then $p\mid x$ and $p\mid y$.

Answer 2

you need to assume that $M$ does not meets a square factor otherwise you can cancel that from both side. That is to say $M=p_1p_2..p_n$ with $p_i$ distinct.

If $p$|$x^2+y^2$ for $p=4k-1$, you can assume that (p,x)=1, otherwise $p|x$ induces $p|y$ then we can cancel $p$ from both side of $M$=$x^2+y^2$

Then we have $x^2 \equiv -y^2 $ $(mod p)$. if $x^2 \equiv a$ then $y^2 \equiv -a$ where $a$ is nonzero, that is $a$ and $-a$ are all quadratic residue of $p$, So $a^{p-1 \over 2} \equiv (-a)^{p-1 \over 2} \equiv 1$ ($mod$ $p$) which will contradict to $p=4k-1$