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Source: 2025 TST T11

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#1 h Mar 10, 2025, 12:00 PM • 1 Y

Y by m4thbl3nd3r

Let $n \geq 4$ . Proof that
$(2^x - 1)(5^x - 1) = y^n$ have no positive integer solution $(x, y)$ .

This post has been edited 1 time. Last edited by EthanWYX2009, Mar 11, 2025, 11:16 AM

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#2 h Mar 10, 2025, 4:22 PM • 3 Y

Y by MS_asdfgzxcvb, mijail, Hoto_Mukai

Some details can be simplified. I will write this in LATEX later.

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DottedCaculator

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DottedCaculator

#3 h Mar 13, 2025, 1:34 PM • 2 Y

Y by mrtheory, pavel kozlov

If $n\mid x$ , then the left hand side is bounded between $(10^{\frac xn}-1)^n$ and $(10^{\frac xn})^n$ , contradiction.

Assume without loss of generality $n$ is prime or $4$ , $6$ , $9$ . We deal with the second case first. As $4\mid 5^x-1$ , we get that $\nu_2(5^x-1)\geq4$ , so $4\mid x$ . Therefore, $3\mid2^x-1$ , so $\nu_3((2^x-1)(5^x-1))\geq4$ , implying $9\mid x$ , so $36\mid x$ , impossible when $n=4,6,9$ .

Now, let $p$ be prime. We show by induction on $p$ that if $n\geq\max(p,4)$ , then every prime $q\leq p$ must divide $y$ and $p\mid x$ . The base cases $p=2,3$ have already been shown. For the inductive step, we already know that every prime $q<p$ must divide $y$ . Then, as $\nu_q((2^x-1)(5^x-1))\geq n\geq p$ and $\nu_q((2^{q-1}-1)(5^{q-1}-1))\leq\max(2,q-1)\leq q$ (easy to show for $q=2,3,5,7$ , then bound for $q\geq11$ ), this implies by Lifting the Exponent that $q^{\left\lceil\frac{n-q}2\right\rceil}\mid x$ . Since $q^{\frac{n-q}2}>n-2$ , this means the largest power of $q$ dividing $x$ is at least $n-1\geq p-1$ , so $p-1\mid x$ . This implies $p\mid y$ . Furthermore, the exponent of $p$ in $(2^x-1)(5^x-1)$ is at most $p-1<n$ as $p\geq5$ , so $p\mid x$ .

Therefore, there are no positive integer solutions when $n\geq4$ .

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#4 h Apr 19, 2025, 7:58 PM

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For $n=2$ it is classic problem, described here https://www.researchgate.net/profile/Laszlo-Szalay-2/publication/264956499_On_the_diophantine_equation_2_n_-13_n_-1x_2/links/588f3f7845851567c9405de3/On-the-diophantine-equation-2-n-13-n-1x-2.pdf

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