3
$\begingroup$

When $x$ is odd, then $x^2+4 \equiv 5 \mod 8$. Can any factor of $x^2+4$ be $\equiv 3\mod 4$? Examples: $$7^2+4=53$$ $$11^2+4= 5^3$$ $$31^2+4= 5 \cdot 193$$ $$47^2+4=2213$$ $$89^2+4=5^2 \cdot 317$$

All of the prime factors above are $\equiv 1 \mod 4$.

Is there no counter example and if so, how to prove this?

  • 0
    If one prime factor of $x^2 + 4$ is congruent to $3 \bmod 4$, then this must also be true of another prime factor, since $3^2 \equiv 1 \bmod 8$. – 2017-02-14
  • 1
    Yes obviously. In fact one of the factors will be $\equiv 3 \mod 8$ and another then has to be $\equiv 7 \mod 8$. But that does not tell us whether it can occur at all. – 2017-02-14
  • 1
    This may or may not be of any help: given $p \equiv 3 \bmod 8$ and $q \equiv 7 \bmod 8$, the Kronecker symbol $$\left(\frac{pq}{4}\right) = -1.$$ – 2017-02-14

2 Answers 2

5

If $p$ is an odd prime divisor of $x^2+4$, then we have $-4 \equiv x^2 \pmod p$

So $1=\left(\frac{-4}{p}\right)=\left(\frac{4}{p}\right) \left(\frac{-1}{p}\right) = \left(\frac{-1}{p}\right)$ which means that $p \equiv 1 \pmod 4$ by the first supplement to quadratic reciprocity.

  • 0
    Right - I totally overlooked that argument; so simple. Thanks! – 2017-02-15
0

$x^2 + 4 \equiv 3 \pmod 4$ is the same as proving $x^2 + 0 \equiv 3 \pmod 4$ since $4 \equiv 0\pmod 4$.

If $x$ is even is easy to see that $x^2 \equiv 0 \pmod 4$ and when $x$ is odd it has the form $2n+1$ then:

$x=(2n+1)^2= 4n^2 + 4n + 1 = 4n(n+1) + 1$ thus $4n(n+1) \equiv 0 \pmod 4$ and $4n(n+1) + 1\equiv 1 \pmod 4$

  • 3
    You have shown that $x^2+4 \not \equiv 3 \pmod 4$ but the question asked if $x^2+4$ could have a factor congruent to $3 \pmod 4$. Clearly if you have one you must have another, but because $3^2 \equiv 1 \pmod 4$ it is not so obvious you can't have one at all. – 2017-02-14
  • 0
    Pretty useless answer. I myself already stated that $x^2 \equiv 1 \mod 8$. – 2017-02-14
  • 0
    I have tried a computer search up to $x=1881$ without finding anything. Baffled ... is it true or false ? – 2017-02-14
  • 0
    @DonaldSplutterwit Intuition suggests no counterexample exists. – 2017-02-14
  • 0
    Surely two factors of the form $3(mod 4)$ must exist for a sufficiently large odd value ? ... please someone crack this! – 2017-02-14
  • 0
    @DonaldSplutterwit From a different direction: I assert that $p = 4k + 3$ as a prime factor does not occur for $k \leq 25$. – 2017-02-14
  • 0
    http://math.stackexchange.com/questions/1472923/prime-divisors-of-x2-1 – 2017-02-14