Search All of the Math Forum:
Views expressed in these public forums are not endorsed by
Drexel University or The Math Forum.



Re: Prove K^2 mod 4 = 1
Posted:
Jan 9, 2013 12:28 AM


One of the procedures for solving this is follows: 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) Hence Prooved.
Message was edited by: johnykeets
Message was edited by: johnykeets



