Solve $n^2 \equiv -1 \mod 4m+3$

Question

The following equation has no solution for $m<1000$ (see computation below) $$n^2 \equiv -1 \mod 4m+3 $$

Question: Is there a solution in general?

sage: for k in range(1000):
....:     for n in range(4*k+3):
....:         if (n**2+1).mod(4*k+3)==0:
....:             print([k,n])
....:
sage:

I see no reason to delete the post. Further, at least some users have valued the other answer, and we very rarely delete upvoted content without good reason. — davidlowryduda, Oct 19 '18 at 13:58
Improved question: Numbers $m$ such that $\sqrt{-1}$ mod $m$ exists — Sebastien Palcoux, Oct 19 '18 at 14:50

score 3 · Answer 1 · edited Nov 13 '18 at 09:31

3

Shorter version of the other answer:

$4m+3$ must have at least one prime factor $p$ of the form $p = 4k+3$. Reducing modulo $p$, the equation reads $n^2\equiv -1\pmod p$, which is impossible by Euler's criterion.

edited Nov 13 '18 at 09:31

Sebastien Palcoux

5,797

answered Oct 17 '18 at 10:46

Arthur

199,419

Sebastien Palcoux · Accepted Answer · 2018-10-19T08:08:35.407

The equation has no solution.

Proof: First assume that $4m+3$ is a prime number $p$. By Euler's criterion we have:
$$ n^2 \equiv -1 \mod p \ \ \Rightarrow \ \ (-1)^{\frac{p-1}{2}} \equiv 1 \mod p.$$ But $\frac{p-1}{2} = 2m+1$, so $-1 = (-1)^{\frac{p-1}{2}} \equiv 1 \mod p$, contradiction with $p$ odd.

Now if $4m+3$ is composite, let $q$ any prime factor. Then $$ n^2 \equiv -1 \mod q$$ but $q$ is odd and by the first part $q \not \equiv 3 \mod 4$, so $q \equiv 1 \mod 4$. It follows that $$3 \equiv 4m+3 \equiv 1^{r} \equiv 1\mod 4$$ with $r$ the number of prime factors of $4m+3$ (with multiplicity), contradiction. $\square$

Solve $n^2 \equiv -1 \mod 4m+3$

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