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.
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?
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
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)