Prove by contradiction If n I\'s an integer and n^2 I\'s even then n I\'s even Solution n^2 is even, but let us assume n is not even. n^2 = 2p n = (2p) LHS is n which is an integer, RHS is (2p) which can be integer only when it is multiple of 2. That is when p is of the form 2k. Then RHS becomes 2k which is even. So if n^2 is even then n is also even..