if $p \not= 2$ and $a^2 \equiv -1 \text{(mod p) }$ then $p \equiv 1 \text{(mod 4)}$

Question

Let $p$ be a prime number which is not equal to 2. Prove that if there exists an integer $a$ such that $a^2 \equiv -1 \text{(mod p) }$ then $p \equiv 1 \text{(mod 4)}$.

My attempt: let $p$ be a prime number not equal to 2. Suppose $p \not \equiv 1 \text{(mod 4)}$. Since $p \not= 2$, p must be odd, therefore $p \equiv 3 \text{(mod 4)}$

We have that $a^2 \equiv -1 \text{(mod p) }$ so $p| a^2 + 1 \implies a^2 + 1 = pm$ for some integer $n$.

$a^2 \equiv 0 \text{(mod 4)} \text{ or } a^2 \equiv 1 \text{(mod 4)} \implies a^2 + 1 \equiv 1 \text{(mod 4)} \text{ or } a^2 + 1 \equiv 2 \text{(mod 4)}$. This means that

$a^2 + 1 \equiv pm \equiv 3m \equiv 1 \text{(mod 4)}$ or

$a^2 + 1 \equiv pn \equiv 3n \equiv 2 \text{(mod 4)}$

But I am not able to reach any contradiction. What am I missing here?

Raise $a^2\equiv-1$ to the $\frac12(p-1)$-th power. – Angina Seng Apr 14 '20 at 19:44 — Angina Seng, Apr 14 '20 at 19:44

score 0 · Accepted Answer · answered Apr 14 '20 at 19:48

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Suppose $p=4n+3$. If $a^2\equiv -1$ for some nonzero residue $a$, then $a^{(4n+2)}=(a^2)^{(2n-1)}\equiv (-1)^{(2n-1)}\equiv -1$. But Fermat's Little Theorem renders a different residue for $a^{(4n+2)}$, can you see that?

answered Apr 14 '20 at 19:48

Oscar Lanzi

48,208

We have not talked about residues in the course I am taking, can you explain it without that? – Carlisle Manson Apr 14 '20 at 20:06
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you should be able to understand the term residue if you understand modular arithmetic – J. W. Tanner Apr 14 '20 at 20:30
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And Fermat's Little Theorem. – Oscar Lanzi Apr 14 '20 at 21:46

if $p \not= 2$ and $a^2 \equiv -1 \text{(mod p) }$ then $p \equiv 1 \text{(mod 4)}$

1 Answers1