Discussion
For which of the following primes \(p\), is the quadratic congruence \(x^{2} \equiv 2 \pmod{p}\) solvable ?
Answer: D
The congruence is solvable for \(p = 23\).
By the second supplement to quadratic reciprocity, 2 is a quadratic residue modulo an odd prime \(p\) exactly when \(p \equiv \pm 1 \pmod 8\). \(23 \equiv 7 \equiv -1 \pmod 8\), so 2 is a residue. Indeed \(5^{2} = 25 \equiv 2 \pmod{23}\).
The others are \(\equiv \pm 3 \pmod 8\): \(5 \equiv 5\), \(13 \equiv 5\) and \(19 \equiv 3\), so 2 is a non-residue and \(x^{2} \equiv 2\) has no solution for them.
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