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Prime Square Divisibility

Problem

Prove that p2minus1 is divisible by 24 for all primes, p > 3.


Solution

For all primes greater than 3, p = 6kplus or minus1.

When p = 6k+1, p2 = 36k2 + 12k + 1 and so p2minus1 = 36k2 + 12k.

Similarly, when p = 6kminus1, p2minus1 = 36k2 minus 12k.

Therefore, p2minus1 = 36k2 plus or minus 12k = 12k(3k plus or minus 1).

If k is odd, 3k plus or minus 1 will be even and so we prove that p2minus1 will always be divisible by 12 times 2 = 24.


What can you say about p3minus1?

What about other powers of p?

Problem ID: 67 (Feb 2002)     Difficulty: 2 Star

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