While learning about the Legendre symbol I came across this fact:
If $x^ 4 \equiv -1 \mod p$ then $p \equiv 1 \mod 8$.
Provided '$p$' is a prime greater than $2$.
I could not prove it. Can someone help me to prove the same?
Thanks.
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Bill Dubuque
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noddy
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$p$ is a prime.? – Inceptio Mar 18 '13 at 08:07
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It is a prime. Else the $p\equiv 2 / 1 \mod 8$ – Inceptio Mar 18 '13 at 08:09
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$p$ prime must be given as assumption. Since $3^4 \equiv -1 \textrm{ mod} 82$, and 82 is not a prime. – Sungjin Kim Mar 18 '13 at 08:17
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p is a prime number – noddy Mar 18 '13 at 08:54
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3You need to specify that $p$ is odd. Note that $1^4 \equiv -1 \pmod{2}$. – Ivan Loh Mar 18 '13 at 09:07
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Let $p$ be a prime. Suppose that $x^4+1\equiv 0 \textrm{ mod }p$ is solvable.
Then $x$ has order 8 in the multiplicative group of $\mathbb{F}_p$.
Since an order of subgroup must divide the group order, and the cyclic group generated by $x$ has order $8$, we have $8\mid p-1$.
Sungjin Kim
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3And since the multiplicative group of $\Bbb F_p$ is itself always cyclic, one has $x^4\equiv1\pmod p$ solvable for an odd prime $p$ if and only if $p\equiv1\pmod8$. (But of course that was not the question.) – Marc van Leeuwen Mar 18 '13 at 08:31