Mathematics
Answer
Given,
p ≥ 5
We know that,
Every prime number greater than 3 is of the form 6k + 1 or 6k + 5.
Let p be of form (6k + 1)
p2 + 2 = (6k + 1)2 + 2
= 36k2 + 1 + 12k + 2
= 36k2 + 3 + 12k
= 3(12k2 + 1 + 4k); which is clearly divisible by 3.
Let p be of the form (6k + 5)
p2 + 2 = (6k + 5)2 + 2
= 36k2 + 25 + 60k + 2
= 36k2 + 60k + 27
= 3(12k2 + 20k + 9); which is clearly divisible by 3.
Hence, proved that for p being prime number and p ≥ 5, p2 + 2 is divisible by 3.
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