China-TST-2025A

We will use the well-known Lifting the Exponent Lemma (LTE), stated as follows: Lifting the Exponent Lemma: Let $p$ be a prime and $m$ a positive integer. Let $a,b$ be integers such that $p\nmid ab$ and $q_p\mid (a-b)$, where $q_p=p$ when $p$ is odd and $q_2=4$ when $p=2$. Then we have $$v_p(a^m-b^m)=v_p(a-b)+v_p(m).$$

Assume there exist positive integers $x,y$ satisfying $(2^x-1)(5^x-1)=y^n$. We first prove the following lemma: Lemma: We have $2\cdot 3\cdot 5\cdot 7\mid x$, and for any prime $p\le n$, $$v_p(x)\ge \lfloor {\log }_{p}n\rfloor .$$

Proof of Lemma: From $2\mid (5^x-1)$ we get $2\mid y$. By LTE: $$\begin{array}{rl}n\le n{v}_2(y)& =v_2((2^x-1)(5^x-1))\\ & =v_2(5-1)+v_2(x).\end{array}$$ Thus $v_2(x)\ge n-2\ge \lfloor {\log }_{2}n\rfloor$. In particular, $4\mid x$, so $3\cdot 5\mid 2^x-1$, hence $3\cdot 5\mid y$. By LTE: $$\begin{array}{rl}n\le n{v}_3(y)& =v_3(((2^2{)}^{\frac{x}{2}}-1)((5^2{)}^{\frac{x}{2}}-1))\\ & =v_3(2^2-1)+v_3(\frac{x}{2})+v_3(5^2-1)+v_3(\frac{x}{2})=2v_3(x)+2;\\ n\le n{v}_5(y)& =v_5(((2^4{)}^{\frac{x}{4}}-1)(5^x-1))\\ & =v_5(2^4-1)+v_5(\frac{x}{4})=v_5(x)+1.\end{array}$$ Thus $v_3(x)\ge (n-2)/2\ge \lfloor {\log }_{3}n\rfloor$, and $v_5(x)\ge n-1\ge \lfloor {\log }_{5}n\rfloor$. In particular, $3\mid x$. Combined with $2\mid x$, we have $6\mid x$, so $7\mid (2^x-1)$, hence $7\mid y$. By LTE: $$\begin{array}{rl}n\le n{v}_7(y)& =v_7(((2^6{)}^{\frac{x}{6}}-1)((5^6{)}^{\frac{x}{6}}-1))\\ & =v_7(2^6-1)+v_7(\frac{x}{6})+v_7(5^6-1)+v_7(\frac{x}{6})=2v_7(x)+2.\end{array}$$ Thus $v_7(x)\ge (n-2)/2\ge \lfloor {\log }_{7}n\rfloor$.

If $n\le 10$, the lemma is already proved. Now assume $n\ge 11$. We proceed by induction on $p$. Let $p$ be a prime with $11\le p\le n$, and assume the lemma holds for all primes $q<p$. By induction hypothesis, for any prime $q\le p-1$, $$v_q(x)\ge \lfloor {\log }_{q}n\rfloor \ge \lfloor {\log }_q(p-1)\rfloor \ge v_q(p-1).$$ Thus $(p-1)\mid x$. By Fermat's Little Theorem, $p\mid (2^{p-1}-1)$ and $p\mid (5^{p-1}-1)$. Hence $p\mid (2^x-1)$ and $p\mid y$. By LTE: $$\begin{array}{rl}n\le n{v}_p(y)& =v_p\left(((2^{p-1}{)}^{\frac{x}{p-1}}-1)((5^{p-1}{)}^{\frac{x}{p-1}}-1)\right)\\ & =v_p(2^{p-1}-1)+v_p\left(\frac{x}{p-1}\right)+v_p(5^{p-1}-1)+v_p\left(\frac{x}{p-1}\right)\\ & =2v_p(x)+v_p(2^{p-1}-1)+v_p(5^{p-1}-1)\\ & <2v_p(x)+{\log }_p(2^{p-1})+{\log }_p(5^{p-1})\\ & <2v_p(x)+p{\log }_p(10).\end{array}$$ Therefore, $$v_p(x)>\frac{n-p{\log }_p(10)}{2}.$$ Since $11\le p\le n$, this implies $v_p(x)>0$, so $v_p(x)\ge 1$. If $n<p^2$, then $v_p(x)\ge 1\ge \lfloor {\log }_{p}n\rfloor$. To complete the induction, we only need to prove for $n\ge p^2\ge 11^2$ that $$\frac{n-p{\log }_p(10)}{2}>{\log }_{p}n.$$ This is equivalent to $$\frac{p^n}{10^p}>n^2.$$ Indeed, using $n\ge p^2\ge 11^2$, we have $$\frac{p^n}{10^p}=\frac{p^p}{10^p}\cdot p^{n-p}>11^{n-\sqrt{n}}>9^{\frac{n}{2}}>n^2.$$ This completes the induction. The lemma is proved.

Proof of the Main Result: By the lemma, for any prime $p\le n$, $$v_p(x)\ge \lfloor {\log }_{p}n\rfloor \ge v_p(n).$$ Thus $n\mid x$. Let $x=n{x}_1$, then $$y^n=(2^x-1)(5^x-1)<(10^{x_1}{)}^n.$$ Hence $10^{x_1}\ge y+1$. Therefore, $$\begin{array}{rl}{y}^n& =(10^{x_1}{)}^n-2^x-5^x+1\\ & \ge (y+1{)}^n-2^x-5^x+1\\ & >y^n+n\cdot y^{n-1}-2\cdot 5^x\\ & \ge y^n+4\cdot y^{\frac{3n}{4}}-2\cdot 5^x.\end{array}$$

This implies $y^{\frac{3n}{4}}<5^x$, i.e., $y^n<5^{\frac{4x}{3}}$. On the other hand, by the lemma we have $x\ge 2\cdot 3\cdot 5\cdot 7$, so $$y^n=(2^x-1)(5^x-1)>(0.9\cdot 2^x)\cdot (0.9\cdot 5^x)>0.8\cdot 5^{\frac{2x}{5}}\cdot 5^x>5^{\frac{4x}{3}},$$ a contradiction. Therefore, the original Diophantine equation has no positive integer solutions. $\square$