Exponential Diophantine Equation Troubles

Hello, fellow Brillianters, how are you all doing?

Recently I came across this question here, and whilst trying to come up with a proof for my solution, I stumbled upon this equation here:

$2*5^{a} - 1 = b^{2}$ , where both a and b are non-negative integers.

I wanted to prove that there are only three pairs of solutions $(a,b)$ for this question; namely, $(0, 1)$ , $(1, 3)$ and $(2, 7)$ .

My first impulse was to try to prove that if any odd integer $b$ , $b > 7$ , is not a solution, then $b + 4$ cannot be a solution as well. I thought that it was sufficient until Calvin Lin came along and showed me that I only proved that $b$ and $b + 4$ cannot be solutions at the same time. Worst part is that he has no idea either of how to prove this.

So here I am, my friends; do you know a way to prove my statement right (or wrong)? I'd appreciate any form of help you can provide me. Thanks!

#NumberTheory #DiophantineEquations #Divisibility #Proof #Stuck

Easy Math Editor

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
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Comments

This is a Ramanujan-Nagell type equation.

According to Wikipedia, a result of Siegel implies that the number of solutions in each case is finite, but not much further is known.

I believe It is unlikely that there is a simply proof of this statement.

Calvin Lin Staff - 6 years, 3 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}`	$\boxed{123}$