IMO 1979 SL 23

Origin: USA

Problem

Find all natural numbers n for which 28 + 211 + 2n is a perfect square.

Solution

It is easily verified that no solutions exist for n \leq8. Let us now assume that n > 8. We note that 28 + 211 + 2n = 28 \cdot (9 + 2n−8). Hence 9 + 2n−8

must also be a square, say 9 + 2n−8 = x2, x \inN, i.e., 2n−8 = x2 −9 = (x−3)(x+3). Thus x−3 and x+3 are both powers of 2, which is possible only for x = 5 and n = 12. Hence, n = 12 is the only solution.