How do i show that the only positive integral solutions to the equation $$10^{x} -5^{x}-2^{x}=y^{2}-1$$ are $(x,y)=(1,2)$.
Tried a lot of things. This looks quite difficult and i am not even sure how to start.
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10$10^x-5^x-2^x+1=5^x(2^x-1)-(2^x-1)=(2^x-1)(5^x-1)=y^2$ possibly helpful. – kingW3 Jul 15 '18 at 14:48
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1the equation has positive integral solutions if and only if $\left( {{2}^{x}}-1 \right)\left( {{5}^{x}}-1 \right)$ is a perfect square – Jul 15 '18 at 15:26
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1No further solution for $x\le 10^5$ – Peter Jul 15 '18 at 17:48
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1A related manuscript. Linked to earlier in this question. – Jyrki Lahtonen Jul 16 '18 at 11:22
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2I suggest you to ask this question: How to proof $\left( {{2}^{x}}-1 \right)\left( {{5}^{x}}-1 \right)$ is not a perfect square number for $x>1$. – user0410 Jul 21 '18 at 18:54
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If you look at the equation $(2^x-1)(5^x-1)=y^2 \mod{3}$, since $5\equiv2\mod{3}$, you get $(2^x-1)^2\equiv y^2\mod{3}$. If I can find a proof based on this, I will post it. – Keith Backman Jul 21 '18 at 21:07