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September 16th, 2017, 07:05 PM   #1
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If P>= 5 is a prime number, then p^2 +2 is composite.

I have seen outlines of the proof, and I tried to fill in the reasons why. Are they right?


If p>=5 is a prime number, then p^2 +2 is composite.

Proof: Suppose p>=5 is a prime number. Then by the quotient remainder theorem, p can be expressed as

6m or 6m + 1 or 6m + 3 or 6m + 4 or 6m + 5 for some integer m.

However, since p is a prime number greater than or equal to 5, it cannot be expressed as a multiple of 2 or 3 greater than 5, because p would then be not prime. Thus p can only be expressed as


6m+1 or 6m+5.


If p = 6m + 1, then by squaring p and adding 2, we get


p^2 + 2 = (6m+1)^2 + 2 = 3(12m^2 + 4m +1)


Thus p^2 + 2 is divisible by a number less than p^2 + 2 and greater than 1, so, p^2 + 2 is composite.

If p = 6m + 5, then by similar reasoning we get


p^2 + 2 = (6m+5)^2 + 2 = 3(12m^2 + 20m +7)


Thus p^2 + 2 is divisible by a number less than p^2 + 2 and greater than 1, so, p^2 + 2 is composite. Hence, in either case p^2 + 2 is composite, which is what we needed to show.
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September 16th, 2017, 10:36 PM   #2
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Yes: $p = 6m \pm 1$ and so $p \equiv \pm 1 \pmod{6}$ and $p^2 \equiv 1 \pmod{6}$ and $p^2 + 2\equiv 3 \pmod{6}$ and thus $p^2+2 = 6k+3$.
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Last edited by v8archie; September 16th, 2017 at 10:39 PM.
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