Posted by **Meg** on Monday, February 14, 2011 at 9:46pm.

Prove that if p is a prime number and p is not equal to 3, then 3 divides p^2 + 2. (Hint: When p is divided by 3, the remainder is either 0,1, or 2. That is, for some integer k, p = 3k or p = 3k + 1 or p = 3k + 2.)

I thought you might do three cases with the three values of p in the hint, plugging them into p^2+2. In two of the cases you get a p^2+2=3a (a some integer) form but for p=3k you do not. Am I approaching this wrong?

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**MathMate**, Monday, February 14, 2011 at 10:13pm
The third case is the simplest, because

3|3k

so you're done!

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**MathMate**, Monday, February 14, 2011 at 10:14pm
On the other hand, since it is a prime ≠ 3, so in fact, the third case does not even arise.

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**SHAKEERA**, Tuesday, February 22, 2011 at 8:11pm
NO

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