Find all pairs yielding a square

Determine all pairs of positive integers $m,n$ such that $2^m + 3^n$ is a perfect square.

Source: Problem Primer for the Olympiad

#NumberTheory #OlympiadMath #MathProblem #Math

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

A brief sketch: Considering modulo $3, m$ must be even. Considering modulo $4, n$ must be even. Thus we have a primitive pythagorean triple, and so we have $2xy = 2^{m/2}$ and $x^2- y^2 = 3^{n/2}$ . This I believe is easily solved and leads to the only solution being $m=4,n=2$ (If I haven't made a mistake).

Peiyush Jain - 7 years, 10 months ago

That's right, well done :)

Tim Vermeulen - 7 years, 10 months ago

Thank you for posting such a nice problem!

Peiyush Jain - 7 years, 10 months ago

@Peiyush Jain – The book I got it from has many of the like, I'd definitely recommend it!

Tim Vermeulen - 7 years, 10 months ago

But also m=0 and n=1 is a solution !!

Mouataz Chadmi - 7 years, 10 months ago

The problem asks for "all pairs of positive integers $m,n$ ". The number $0$ is not positive.

Jimmy Kariznov - 7 years, 10 months ago

This problem is from Indian National Mathematical Olympiad

ARGHYA DATTA - 7 years, 10 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$