0
$\begingroup$

Consider a number on the form, $N=4^x + k^2$ for integer $x \geq 1$ and odd integer $k \geq 1$, so $N=4^x + (2y+1)^2 = 4^x + 4y^2 + 4y+ 1$ for integer $y \geq 0$. Clearly, $N \equiv 1 \pmod 4$.

Are all prime factors of $N$ on the form $4z+1$?

It is clearly the case when $N$ is prime, and appears to be like that for all other values of $x$ and $y$ that I have tried.

As I understand it Fermat's theorem on sums on two squares only says that $N \equiv 1 \pmod 4$ as the number is expressed as a sum of two squares (although it is also clear from above). The Brahmagupta–Fibonacci identity says that there is an even number of prime factors on the form $4z+3$ when $N$ can be written as a sum of two squares. I know that $n^2+1$ only has prime factors on the form $4z+1$, so for $k=1$ it holds.

However, I do not manage to generalize this to the general case. Nor am I sure that it actually holds (which is the major issue here, I want to state that) although simulations have yet to find a case where it does not hold.

  • 1
    I think most of what you've written after "As I understand" is a little confused. For example, the Brahmagupta-Fibonacci identity says no such thing. What you actually get is that if $a^2+b^2=p_1^{\alpha_1}\cdots p_k^{\alpha_k}$ is the prime factorization of the sum of two squares, then if $p_i\equiv 3\pmod{4}$ then $\alpha_i$ is even. It says nothing about the number of prime factors of the form $4z+3$. – 2017-01-09
  • 0
    You are correct. A bit to quick when trying to read up. Not my area really. Should I edit the question to remove the confused parts? (Not sure what the preferred behaviour is.) – 2017-01-09

1 Answers 1

1

More generally, the only prime factors $p$ of $a^2+b^2$ are primes $p\equiv 1\pmod{4}$ and divisors of $\gcd(a,b)$.

This is because if $p\mid a^2+b^2$ and $p\not \mid a$, then $au\equiv 1\pmod{p}$ for some $u$ and $(bu)^2\equiv -a^2u^2\equiv -1\pmod {p}$, and thus $-1$ is a square modulo $p$.

In your case, $a=2^x$ and $b=k$ and $\gcd(a,b)=1$ since $k$ is odd, so the only prime divisors satisfy $p\equiv 1\pmod 4$.