Fairly easy Pick m=7^n, by LTE it = 4^7^n+2^7^n+1=8^7^n-1:2^7^n-1[7^(n+1)], done by LTE

This is also Iranian third round NT exam Finals problem 3.

Can you explain the solution? I solved it by induction

We know that for all $m$ $$7|3^m+5^m-1$$is true. SInce 7 is prime, we can use Fermat's little theorem and observe that if $m=7^k$, the above condition is right.Seeing this, we can check that $n=1, m=7^n=7, n=2, m=7^n=49$ works. So it motivates us to pick $m=7^n$ Now, we will prove $n \le v_7(3^m+5^m-1)$ $$3^{7^n}+5^{7^n}-1=3^{7^n}+(4+1)^{7^n}-1=3^{7^n}+4^{7^n}+ \sum_{k=0}^{n-1} \binom{n}{k} 4^k-1$$ Note that $v_7(3^{7^n}+4^{7^n})=v_7(3+4)+v_7(7^n)=1+n$ Therefore it is enough to prove $$v_7 \left( \sum_{k=0}^{7^n-1} \binom{7^n}{k} 4^k-1 \right) \ge n$$ And: $$ \sum_{k=0}^{7^n-1} \binom{7^n}{k} 4^k-1 = \binom{7^n}{1}4^1+\binom{7^n}{2}4^2+...+\binom{7^n}{7^n-1}4^{7^n-1}$$It is easy to see $v_7 \left( \binom{7^n}{1}4^1+\binom{7^n}{2}4^2+...+\binom{7^n}{7^n-1}4^{7^n-1} \right) \ge n$ Hence $v_7(3^m+5^m+1) \ge n$ meaning $7^n | 3^m +5^m -1$

For $n=1$ and $n=2,$ $m=1$ and $m=7$ satisfies given condition,respectively. So it motivates us to pick $m=7^{n-1}$ Note that $$v_7(3^m+4^m)= v_7(3+4)+v_7(m)=n \iff 7^n \mid 3^m+4^m$$$$v_7(2^m+5^m)= v_7(2+5)+v_7(m)=n \iff 7^n \mid 2^m+5^m$$ Thus we just need to show $$ 7^n \mid 2^m+4^m+1 \iff 7^n \mid \frac{2^{3m}-1}{2^m-1}$$Since $3 \nmid m \implies 7 \nmid 2^m-1$.Hence it suffices to show that $$7^n \mid 8^m-1 \iff v_7(8^m-1) \ge n \iff v_7(8-1)+v_7(m) \ge n \iff n \ge n$$ So we are done