Problem
NT-B2-M02-P010 A Prime Divisor of \(a^2+1\)
#10
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Let an odd prime \(p\mid a^2+1\). Prove that \(p\equiv1\pmod4\).
From \(a^2\equiv-1\pmod p\), obtain an element of order \(4\).
Since \(p\mid a^2+1\), we have \(a^2\equiv-1\pmod p\), and \(p\nmid a\). Then \(a^4\equiv1\), but \(a^2\not\equiv1\). Hence the order of \(a\) modulo \(p\) is \(4\). The order of an element divides \(p-1\), so \(4\mid p-1\), meaning \(p\equiv1\pmod4\).
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