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In this answer of MO it is showed that $$\frac{a^m-1}{a-1}=2b^2$$ has no solution for $m>2$.

I would like to check if using the same argument, it can be stated that $$\frac{a^m-1}{a-1}=\frac{2b^2}{c}$$ where $c,b$ are odd, has no solution for $m>2$. And, if possible, digging a bit the original argument to have a better understanding of it, even if it is not valid for the new stated equation.

Thanks!

Juan Moreno
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    $$\frac{3^4-1}{3-1}=\frac{2\cdot 10^2}{5}$$ hence the second equation has a solution for $m>2$. – Peter Oct 15 '22 at 10:07
  • @Peter thanks for your counterexample. However, my fault, as I forgot to state $c,b$ odd. If you do not mind, I edit the OP to include this forgotten restriction. – Juan Moreno Oct 15 '22 at 10:58
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    @Blue many thanks for your edits, now the post looks much nicer! – Juan Moreno Oct 15 '22 at 10:59
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    There is still a solution : $$\frac{5^6-1}{5-1}=\frac{2\cdot 651^2}{217}$$ – Peter Oct 15 '22 at 11:28
  • @Peter thanks! In that case, I would like to understand why in these cases there are solutions, and there is no solution for $2b^2$, could you please share how did you derive the counterexamples? – Juan Moreno Oct 15 '22 at 11:49
  • The counterexample was easy to find, we only need that the left side is divisible by $2$ , but not by $4$ , and since my expression happened to be divisible by $9$ , I could even avoid the trivial case $c=b$. Why this does not work for $2b^2$ ? I guess the proof you mentioned is quite complicated , I doubt that this can be easily understood. – Peter Oct 15 '22 at 11:56

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