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?


  • $\begingroup$ $p$ is a prime.? $\endgroup$ – Inceptio Mar 18 '13 at 8:07
  • $\begingroup$ It is a prime. Else the $p\equiv 2 / 1 \mod 8$ $\endgroup$ – Inceptio Mar 18 '13 at 8:09
  • $\begingroup$ $p$ prime must be given as assumption. Since $3^4 \equiv -1 \textrm{ mod} 82$, and 82 is not a prime. $\endgroup$ – Sungjin Kim Mar 18 '13 at 8:17
  • $\begingroup$ p is a prime number $\endgroup$ – noddy Mar 18 '13 at 8:54
  • 3
    $\begingroup$ You need to specify that $p$ is odd. Note that $1^4 \equiv -1 \pmod{2}$. $\endgroup$ – Ivan Loh Mar 18 '13 at 9:07

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 $8\mid p-1$.

  • 3
    $\begingroup$ And 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.) $\endgroup$ – Marc van Leeuwen Mar 18 '13 at 8:31

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