Let \(p\) be a prime such that there exists an integer \(x\) satisfying \[x^4\equiv -1\pmod{p}\]

Prove that \(p\equiv 1\pmod{8}\)

*Source: Classic*

Let \(p\) be a prime such that there exists an integer \(x\) satisfying \[x^4\equiv -1\pmod{p}\]

Prove that \(p\equiv 1\pmod{8}\)

*Source: Classic*

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## Comments

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TopNewestIf \(x\) satisfies \(x^4\equiv -1 \pmod p\), then we claim that \(8\) is the order of \(x\), this will imply \(8|p-1\).

Suppose the order is not \(8\), then by properties of order it must be a divisor of \(8\). This is absurd since either \(x\equiv -1\) or \(x^2\equiv -1\) would contradict \(x^4\equiv -1 \pmod p\).

This can clearly be generalized for any powers of 2. – Xuming Liang · 1 year, 10 months ago

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