Even/odd proofs: Practice problems Solutions
Since, by assumption, s = n2 and t = m2 are odd, the integers n and m must be odd as well (by Problem 2). Hence n = 2k +1 and m = 2l +1 for some k;l 2Z, by the de nition of an odd integer. Since the sum of two odd numbers is even (by Problem 1), s+t = p2 is even. Hence p, must be even as well (by Problem 2).
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