Thread: Prove K^2 mod 4 = 1 ?

This is my first post

I have been working on this for far too long

Can someone let me know if this is correct?

"k = 2n + 1. Then k^2 (is congruent to) 4n^2 + 4n + 1 (is congruent to) 1 mod 4"

Should I add more to it? Am I at least on the right track?

Hey brianjasperman.

What you did is correct but if your teacher is picky you should probably write 4n^2 + 4n + 1 = 4(n^2 + n) + 1 = 4m + 1 just to add that extra clarity to the proof.

So I would prove the equation by saying:
"k = 2n + 1. Then 4n^2 + 4n + 1 = 4(n^2 + n) + 1 = 4m + 1"

OR

"k = 2n + 1. Then k^2 (is congruent to) 4n^2 + 4n + 1 (is congruent to) 1 mod 4, and 4n^2 + 4n + 1 = 4(n^2 + n) + 1 = 4m + 1"

Thanks for the help

Basically 4m + 1 where m is an integer implies congruence to 1 (MOD 4) and you're done.

Thank you very very much for your help chiro, but I'm not understanding what your saying.

Can you show me what the entire proof would be (for Christmas?).

Let k = 2n + 1 where n is an integer (in Z).

Then k^2 = (2n + 1)^2 = 4n^2 + 4n + 1 = 4(n^2 + n) + 1 = 4m + 1 for some integer m (m in Z).

By the definition of congruence something is congruent z = r (MOD q) iff z = pq + r where 0 <= r < q for some integers p, r, q and z.

By letting z = k^2, p = m, r = 1, and q = 4 and noting 0 <= 1 < 4 and all variables are integers, we get our condition that

k^2 = (1 MOD 4)