BMO 2017

We prove this by induction on $n$. The case $n=1$ is indeed trivial for $m=1$.

Assume that the statement of the problem holds true for $n$, and we have $3^m+5^m-1=7^n$ for some positive integer $l$ which is not divisible by $7$ (if not we are done). Since $3^6\equiv 1(\mathrm{mod}7)$ and $5^6\equiv 1(\mathrm{mod}7)$ we conclude that, $$3^{6\cdot 7^{n-1}}\equiv 1(\mathrm{mod}{7}^n),5^{6\cdot 7^{n-1}}\equiv 1(\mathrm{mod}{7}^n).$$ Since $$v_7(3^{6\cdot 7^{n-1}}-1)=v_7(3^6-1)+v_7(7^{n-1})=n\text{and}{v}_7(5^{6\cdot 7^{n-1}}-1)=v_7(5^6-1)+v_7(7^{n-1})=n.$$ Thus we can say that: $3^{6\cdot 7^{n-1}}=1+7^{n}r$, $5^{6\cdot 7^{n-1}}=1+7^{n}s$ for some positive integers $r,s$. We find the remainder of $r,s$ modulo $7$. Note that: $$\frac{y^{7^k}-1}{7^{k+1}}=\frac{y-1}{7}\cdot \frac{1+y+\cdots +y^6}{7}\cdots \frac{1+y^{7^{k-1}}+\cdots +y^{6\cdot 7^{k-1}}}{7}(\ast )$$ We use the above identity for $y=3^6,5^6$. Note that in both cases $y\equiv 1(\mathrm{mod}7)$. Now we use the following lemma

Lemma. Let $p$ be an odd prime such that $p\mid a-1$ then $\frac{a^p-1}{a-1}\equiv p(\mathrm{mod}{p}^2)$.

Proof. Take $a-1=b$, then $p\mid b$ now $$\frac{a^p-1}{a-1}=\frac{(b+1{)}^p-1}{b}=b^{p-1}+\cdots +\left(\frac{p}{2}\right)b+p\equiv p(\mathrm{mod}{p}^2)$$ since all the binomial coefficients are divisible by $p$. So our proof is complete. ■

Then by use of the lemma repeatedly we find that all the terms of the above identity (*) except the first term are congruent to $1$ modulo $7$. Thus we can find that: $$\frac{y^{7^k}-1}{7^{k+1}}\equiv \frac{y-1}{7}(\mathrm{mod}7)$$ Since $$\frac{3^6-1}{7}=104\equiv -1,\frac{5^6-1}{7}=2232\equiv -1(\mathrm{mod}7)$$ we find that $r\equiv s\equiv -1(\mathrm{mod}7)$, and by use of binomial theorem, we can easily find that $$3^{6t\cdot 7^{n-1}}\equiv 1+7^{n}r(\mathrm{mod}{7}^{n+1}),5^{6t\cdot 7^{n-1}}\equiv 1+7^{n}s(\mathrm{mod}{7}^{n+1})$$ for all positive integers $t$.

Now take $m+6t\cdot 7^{n-1}$ instead of $m$, (while we will specify the number $t$ later) we can find that: $$3^{m+6t\cdot 7^{n-1}}+5^{m+6t\cdot 7^{n-1}}-1=3^m\cdot 3^{6t\cdot 7^{n-1}}+5^m\cdot 5^{6t\cdot 7^{n-1}}-1$$ Taking modulo $7^{n+1}$ we can find that, the above expression is reduced to $$\begin{gathered} 3^m(1+7^{n}r)+5^m(1+7^{n}s)-1\equiv 3^m+5^m-1+5^m\cdot 7^{n}s+3^m\cdot 7^{n}r \\ \equiv 7^n(1+(5^{m}s+3^{m}r)t)(\mathrm{mod}{7}^{n+1}) \end{gathered}$$ Now, the problem reduces to finding a positive integer $t$ such that $$1+(5^{m}s+3^{m}r)t\equiv 0(\mathrm{mod}7)$$ since $$5^{m}s+3^{m}r\equiv -5^m-3^m\equiv 1(\mathrm{mod}7).$$ Since $3^m+5^m-1\equiv 0(\mathrm{mod}7)$ whence, we find that $\gcd (5^{m}s+3^{m}r,7)=1$. Thus such integer $t$ exists, so we are done!