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# For some integer p, p^2 - 5 is NEVER divisible by 3. WHY?

147 Points
11 years ago

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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Regards,

Sanjeev K Saxena
4 Points
11 years ago

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.