For which value of n is 274 + 22058 + 22n a perfect square?

A sum of three terms forms a perfect square when it matches the identity (a + b)2 = a2 + 2ab + b2 — two squared terms plus twice their product as the middle (cross) term.

1. Match the first term to a square: 274 = (237)2, so take a = 237.

2. The third term is already a perfect square too: 22n = (2n)2, so take b = 2n.

3. For the expression to equal (a + b)2, the remaining term must be the middle term 2ab: 2 · 237 · 2n = 22058, i.e. 2n+38 = 22058.

4. Equate the exponents: n + 38 = 2058, so n = 2020.

Cross-check the only other pairing: treating 22058 itself as the second square (b = 21029) would force the middle term 2ab = 2 · 237 · 21029 = 21067 to equal 22n; but 1067 is odd, and 2n is always even, so this pairing is impossible — confirming n = 2020 is the unique solution.

Hence n = 2020, and the expression equals (237 + 22020)2 — a genuine perfect square.