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`        For some integer p, p^2 - 5 is NEVER divisible by 3. WHY?`
7 years ago

147 Points
```										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
Please feel free to post as many doubts on our discussion forum as you can.If you find any question Difficult to understand - post it here and we will get you the answer and detailed  solution very  quickly. We are all IITians and here to help you in your IIT JEE preparation.All the best. Regards,Askiitians ExpertsBadiuddin
```
7 years ago
Sanjeev K Saxena
4 Points
```										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.
```
7 years ago
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