if p is prime then p^2 = 1 mod 24

Problem:
Prove that for \( p > 3 \) prime, \( p^2 \equiv 1 \pmod{24} \).

Solution:
Consider \( p^2 - 1 = (p-1) (p+1) \).
Since \(p > 3\), we know that either \(p-1\) or \( p+1 \) is divisible by 3. Both are divisible by 2. Also, one of them is divisible by 4, since p is either \( -1 \) or \( 1 \pmod 4 \). Hence \( p^2-1 \) is divisible by \( 2 \times 3 \times 4 = 24 \).

QED

Comments

641 | 2^32 + 1

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12 coins

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sum of 3 out of 5 is divisible by 3

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Maths by Chance

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