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.