Solution to puzzle 155: Sum of two powers is a square?

Is 2ⁿ + 3ⁿ (where n is an integer) ever the square of a rational number?

We will first show that 2ⁿ + 3ⁿ is never a perfect square if n is a positive integer. The proof for negative n will then follow easily. We will use a technique that is often useful in number theory problems: consider the expression to various moduli.

Working modulo 3, we have 2ⁿ + 3ⁿ (−1)ⁿ.
Since all squares mod 3 are equal to 0 or 1, we conclude that n is even.
(We could arrive at the same conclusion working modulo 4.)

Now, letting n = 2m, and working modulo 5, we have 4^m + 9^m (−1)^m + (−1)^m ±2.
All squares mod 5 are equal to 0, 1, or 4; never ±2.
Thus, 2ⁿ + 3ⁿ is never a perfect square if n is a positive integer.

Now we consider negative n.
Let n = −m, and consider 2ⁿ + 3ⁿ = 1/2^m + 1/3^m = (2^m + 3^m)/6^m.
If d | 2^m + 3^m and d | 6^m, then d | 2^m(2^m + 3^m) − 6^m = 2^2m.
Similarly d | 3^2m.
But 2^2m and 3^2m are relatively prime, and so d = 1.
That is, 2^m + 3^m and 6^m are relatively prime.
Hence, if (2^m + 3^m)/6^m is a square, both 2^m + 3^m and 6^m must be square.
But we have already seen that 2^m + 3^m is never a square when m > 0, and so (2^m + 3^m)/6^m is never a square.

Finally, it is clear that 2ⁿ + 3ⁿ is not a square when n = 0.

Therefore, 2ⁿ + 3ⁿ is never the square of a rational number.

For which positive integers x, y is 2^x + 3^y a perfect square?

Source: Online Encyclopedia of Integer Sequences: A114705; see comments