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

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?

• You may want your $4k-1$ primes to be distinct or your squares to be positive, as $3 \times 3 \times 7 \times 7 = 441=21^2+0^2$ – Henry Jun 15 '18 at 12:10
• Uhm... What about $9^2+32^2=5*13*17$? You can infact verify that numbers which are expressible as sum of two squares are closed under multiplication, so I don't see any obstacle to have $4k-1$ primes (even distinct). Infact notice that primes = 1 mod 4 are always expressible as sum of two squares. – frame95 Jun 15 '18 at 12:19
• Do you mean the set of primes cannot have size (4k-1) or all primes in the set are not equivalent to 4k-1 $\pmod p$? – David Diaz Jun 15 '18 at 12:20

Hint:

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

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 $p=4k-1$