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For some integer p, p^2 - 5 is NEVER divisible by 3. WHY?
Dear Sanjeev
Case : p is prime to 3
then p2 -1 =multiple of 3=3k
so p2-5 = p2-1 -3-1
= 3k -3 -1
=3(k-1) -1
-1 is not divisible by 3
Case 2 : p is not prime to 3
then p=multiple of 3
p2 = multiple of 3 =3k
so p2-5 = p2-6 +1
= 3k -6 +1
=3(k-2) +1
+1 is not divisible by 3
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Thank you so much for making me believe that there is only one approach and explanation to the problem in hand. Please congratulate me for an exactly similar explanation that I had before hand.
There are only two resolutions for p, either it’s PRIME to 3 or NOT. In the first case we can safely take p^2 - 1 as some multiple of 3, say 3 m (m is a positive integer). Now, p^2 - 5 = p^2 - 1 - 3 - 1 = 3 m - 3 - 1 = 3 (m - 1) - 1, and 3 CANNOT divide -1. In the second case, when p is not prime to 3, then p is a multiple of 3, let’s again say p^2 = 3 m so that p^2 - 5 = p^2 - 6 + 1= 3 m - 6 + 1 = 3 (m - 2) + 1, and again, 3 CANNOT divide +1 either.
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