Dr Reasoner is not correct.

It is reasonably straightforward to find more than one m and n that makes 2^m + 3ⁿ a perfect square:

e.g.

m = 0 and n= 1 gives a value of 4

m =3 and n = 0 gives a value of 9

m = 4 and n = 2 gives a value of 25

Therefore, there cannot be only one pair of integers m and n that make a perfect square.

Note

Dr Reasoner has realised that if the values of m and n have to be positive integers, then her claim is correct. Here is a proof of this, based on work sent in to us by Aoma in Y11. The proof uses modular arithmetic.

Suppose that 2^m + 3ⁿ = q². (*)

Note that as 2^m must be even and 3ⁿ then q² must be odd.

Note too that any odd square has the form q²= (2k + 1)² = 4k² + 4k + 1 = 4k(k + 1) + 1.

Working in mod 3 arithmetic, if k is

a multiple of 3, then q² = 1 (mod 3)
one less than a multiple of 3, then q²= 1 (mod 3)
one more than a multiple of 3, then q² = 4 x 1 x 2 + 1 (mod 3) = 9 (mod 3) = 0 (mod 3)

Hence q² can only equal 1 or 0 in mod 3 arithmetic.

Let’s look at (*) in mod 3 arithmetic:

q² = (3 – 1)^m + 3ⁿ = (-1)^m(mod 3),

so either (-1)^m = 0 (mod 3) , which is impossible, or (-1)^m = 1 (mod 3), making m even.

Let m = 2M.

Let’s look at (*) in mod 4 arithmetic:

q² = (3 – 1)^2M + (4 – 1)ⁿ = (2)^2M + (-1)ⁿ= 4^M + (-1)ⁿ = (-1)ⁿ(mod 4)

so either (-1)ⁿ = 0 (mod 4) , which is impossible, or (-1)ⁿ = 1 (mod 4), making n even.

Let n = 2N.

Now we know that we need 2^2M + 3^2N = q², where 2M and 2N are both positive even integers (so M and N are positive integers)

Rearranging, 3^2N = q² – 2^2M = (q + 2^M)(q – 2^M)

Either

q + 2^M and q – 2^M are both powers of 3 which are divisible by 3, say 3^r and 3^{2N – r}(r > 0, r >2N-r)) , in which case by subtracting, we get

3(3^r-1 – 3^2N-r-1) = 2(2^M) = 2^M+1

This is impossible, because 3 is a divisor of the left hand side but not the right hand side.

Or

q + 2^M = 3^2N and q – 2^M = 1, in which case by subtracting, we get

3^2N – 1 = 2(2^M), hence (3^N + 1) (3^N – 1) = 2^M+1.

If this is so, then we need a factor pair of 2^M+1 where the factors differ by just 2. All factor pairs for 2^M+1 must be powers of 2 and the only powers of 2 that differ by exactly 2 are 2 and 4. This happens when N = 1.

Hence we would have 2^M+1 = 4 x 2 = 8, so M= 3 – 1 = 2.

Therefore, with m = 2M = 4 and n = 2N = 2, we have the only pair of positive integers which produce a square number from this expression.

Quick check: m = 4, n = 2

2^m + 3ⁿ = 16 + 9 = 25, which is indeed a square number.

Challenge 324: Reasoner's Return III