Let n be a positive integer and suppose that p=2k+1 is an odd prime divisor of n^2+1.
CONG implies congruent to.
1)Show that n^2 CONG -1 mod p
I've done this by saying, because p|n^2+1 --> n^2+1 CONG mod p hence we arrive at the desired result by moving 1.
2) Using Fermat Little Theorem show that (n^2)^k CONG 1 mod p
I've done this by using the standard version of Fermat's little Theorem,
i.e a^p-1 CONG 1 mod p
Using p=2k+1 --> p-1=2k and setting a=n
We get, n^(2k) CONG 1 mod p and fixing up n^2k to (n^2)^k CONG 1 mod p we arrive at the desired result.
3)Deduce from the preceding two parts, show that p CONG 1 mod 4.
Now here is where I'm in doubt, I'm confident with my previous answers(I don't have the solutions to these btw).
Setting p=4, causes problems with p=2k+1 --> 3=2k .. k ?
So this is where I need your guidance.