Recent content by blairebear

Or can I just say p would not be prime since p would be divisible by 3? But then that also holds of 6q, 6q+2 and 6q+4. The only possible primes are 6q+1 and 6q+5 which I proved are composites. OK I think I am done.

Homework Statement If p>=5 is prime, prove that p^2 +2 is composite. Homework Equations If we took p and divided it by 6 we would get remainder possibilities of 0, 1, 2,3,4,5 The Attempt at a Solution p=6q p^2=36q^2 P^2=6(r) P^2+2=6r+2=2(3r+1) composite p=6q+1...

OK, I am going to say this is not true. When a=3 then a^2 is in the form 3k. so I have disproved it I think. However the next part says if 2 or 3 do not divide a. That eliminates every third number. I also read my proof wrong and it is either 3k or 3k+1 not 3k+2 so eliminating all the times...

Ok, I just realized that 3k and 3k+2 are not necessarily even and 3k+1 is not necessarily odd. So now I am even more confused

That is not what I mean to say. based on before I know that 8 divides a^2-1 and 3 divides a^2-1. Can I then say that since both 3 and 8 divide a^2-1 then 24 divides a^2-1

Homework Statement prove that if 2 does not divide a then 24 divides a^2-1 Homework Equations I know that if 2 does not divide a then a is odd. I proved that the square for all odd integers are of the form 8K+1 I also proved the square of any integer is either in the form...