1. Prove that for all natural numbers n ≥ 3 there exist odd natural numbers xn, yn such that 7x 2 n + y 2 n = 2n. Solution: For n = 3 we have x3 = y3 = 1. Now suppose that for a given natural number n we have odd natural numbers xn, yn such that 7x 2 n + y 2 n = 2n; we shall exhibit a pair (X, Y ) such that 7X2 + Y 2 = 2n+1. In fact, 7 xn ± yn 2 2 + 7xn ∓ yn 2 2 = 2(7x 2 n + y 2 n ) = 2n+1 . One of (xn +yn)/2 and |xn −yn|/2 is odd (as their sum is the larger of xn and yn, which is odd), giving the desired pair.