Solving $2 \times 5^n - 1 = m^2$ in natural numbers

Asked Jun 04 '17 at 02:34

Active Jun 04 '17 at 02:34

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This problem isn't of any particular importance in itself, but came up for me randomly, and though I couldn't crack it, I thought learning how would likely be instructive in improving my number theory skills.

Is it possible to completely classify the solutions to $2 \times 5^n - 1 = m^2$ in natural numbers (i.e., the values of $n$ for which the left-hand expression is indeed a square)? Those $n \in \{0, 1, 2\}$ clearly work, and after that most don't, but are these indeed all the solutions, and if so (or if whatever alternative holds), how does one see so?

asked Jun 04 '17 at 02:34

Sridhar Ramesh

1,643

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https://math.stackexchange.com/questions/124565/constructive-proof-need-to-know-the-solutions-of-the-equations – Brevan Ellefsen Jun 04 '17 at 02:57
Ah, thank you for the links. – Sridhar Ramesh Jun 04 '17 at 02:59
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No problem; thank approach0.xyz. Also, let it be noted: this problem does have some importance in itself, because it is hard to solve, relates to factoring over Gaussian Integers, could help with related diophantine equations, and, most importantly, is a genuine mathematical inquiry, which are vital to discovering new mathematics – Brevan Ellefsen Jun 04 '17 at 03:03
What you could try doing is considering a solution to $2\cdot 5^n=m^{2}+1$, proceed to factor both sides in terms of primes in $\mathbb{Z}(\sqrt{-1})$. Another user linked to this though so this should be ignored. – Dave huff Jun 04 '17 at 03:04
@Davehuff did you see the duplicate link? – Brevan Ellefsen Jun 04 '17 at 03:05
@BrevanEllefsen I did not, I just looked at it now. – Dave huff Jun 04 '17 at 03:05

Solving $2 \times 5^n - 1 = m^2$ in natural numbers

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