Re: 2(p^2)+1 = 0 mod 3 for any prime p != 3

Lewis McCarthy (lmccarth@cs.umass.edu)
Wed, 25 Mar 1998 15:02:57 -0500

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Hamdi Tounsi writes:
> can someone give me a proof for that ?

Actually we can prove a stronger, though less interesting, result:
        2m^2 + 1 = 0 (mod 3) iff m != 0 (mod 3)

If m != 0 (mod 3) then m has the form 3k + 1 or 3k + 2, for some k.
2(3k + 1)^2 + 1 = 2(9k^2 + 6k + 1) + 1 = 18k^2 + 12k + 3 = 0 (mod 3)
2(3k + 2)^2 + 1 = 2(9k^2 + 12k + 4) + 1 = 18k^2 + 24k + 9 = 0 (mod 3)

If m = 0 (mod 3) then m has the form 3k for some k.
2(3k)^2 + 1 = 18k^2 + 1 = 1 (mod 3)

-- 
Lewis    http://www.cs.umass.edu/~lmccarth/
"You're talking to a tourist whose every move's among the purest;
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