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Suppose $p$ is an odd prime and there exists some integer $x$ such that $x^2 \equiv -1 \pmod{p}$. Prove that $p \equiv 1 \pmod{4}$. Use Fermat's Theorem.

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closed as off-topic by ThePortakal, Henrik, Foobaz John, Vidyanshu Mishra, kingW3 Oct 1 '17 at 19:17

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$\mathbb{F}_p^{\times}$ is cyclic of order $p-1$ (even). So by Fermat-Lagrange result we have that for every elements : $x^{p-1}\equiv 1 \pmod p$.

Morover with the hypothesis we have the fact that $x$ has order $4$. Indeed $x^4\equiv 1 \pmod p$. So $4 \mid p-1$.

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Let's call the order of $x$ to be the smallest positive integer power, $k$ so that $x^k \equiv 1 \mod p$. We'll denote the order of $x$ as $|x|$.... (assuming some a power exists; which it won't if $x$ is not relatively prime to $p$.)

Clearly $x^{k|x| + j; 0 \le j < |x|} = (x^{|x|})^k*x^j \equiv 1^k*x^j\equiv x^j \mod p$ and furthermore $x^{m} \equiv 1 \mod p \iff m \equiv 0 \mod |x|$

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Okay we know 1) $x^2 \equiv -1\mod p$.

And 2) Fermat's Theorem: $x^{p-1} \equiv 1 \mod p$.

For 1) It is easy to verify $|x| = 4$. [$*$].

So 2) implies $p-1 \equiv 0 \mod 4$.

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[$*$]If the order of $x$ were $1$ then $x\equiv 1\mod p \implies x^2 \equiv 1\not \equiv -1 \mod p$.
If the order of $x$ were $3$ then $1 \equiv x^3 = x*x^2 \equiv -x \mod p\implies x\equiv -1 \mod p \implies x^2 \equiv 1 \mod p$. And, of course $x^4 \equiv 1 \mod p$. [Note; this assumes $p > 2$ so $1 \not \equiv -1 \mod p$]

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Your answer is in theorem $2.12$. Just exclude $p=2$ as you said odd primes.

  • Source: Niven, Zukerman, Montgomery- An introduction to theory of numbers.

Theorem 2.12

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