0
$\begingroup$

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?

$\endgroup$
  • 3
    $\begingroup$ You may want your $4k-1$ primes to be distinct or your squares to be positive, as $3 \times 3 \times 7 \times 7 = 441=21^2+0^2$ $\endgroup$ – Henry Jun 15 '18 at 12:10
  • 4
    $\begingroup$ 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. $\endgroup$ – frame95 Jun 15 '18 at 12:19
  • 1
    $\begingroup$ Do you mean the set of primes cannot have size (4k-1) or all primes in the set are not equivalent to 4k-1 $\pmod p$? $\endgroup$ – David Diaz Jun 15 '18 at 12:20
1
$\begingroup$

Hint:

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

$\endgroup$
0
$\begingroup$

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 $p=4k-1$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.