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 September 16th, 2017, 07:05 PM #1 Newbie   Joined: Sep 2017 From: San Diego Posts: 8 Thanks: 0 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. September 16th, 2017, 10:36 PM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,685 Thanks: 2665 Math Focus: Mainly analysis and algebra 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 $p^2+2 = 6k+3$. Thanks from Shadow89 Last edited by v8archie; September 16th, 2017 at 10:39 PM. Tags composite, number, p>, p>, prime Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post mobel Number Theory 24 October 30th, 2014 09:18 AM agustin975 Number Theory 8 March 8th, 2013 09:04 AM ershi Number Theory 113 August 18th, 2012 04:45 AM Mighty Mouse Jr Algebra 6 May 11th, 2010 09:08 AM Geir Number Theory 1 July 11th, 2009 04:57 AM

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