$f(n) = k\times 124^n +1$ Let $f(n) = 1 + k \times124^n$ where $k$ is a positive integer. Assume there exists distinct integers $m,n \ge 2$ such that either $m,n \mid f(i)$ for all $i$. Prove there exists integer $x \ge 2$ such that $x \mid f(n)$ for all $n$.

I tried some small cases like $k=1$, but I didn't see any patterns. For $k=1$, I noticed that $f(1) = 125 = 5^3$ and that $f(2)$ was prime. Could that mean that one of the numbers must be $f(2)?$ and what could $r$ be. I also noticed that $f(1) \mid f(3)$ if $k=1$. However, I couldn't see this in the $k=2$ case, at least for the terms I could compute. However, for $k=2$ I noticed divisibility by $3$. I tried finding a contriadiction and find possible $x$, but I can't seem to find any.
 A: Without loss of generality, we may assume that $m, n$ are prime numbers, which are both prime to $k$ and $124$.
Now consider $f(1)$, $f(2)$ and $f(3)$. We may assume that two of them are divisible by $m$, say $f(a), f(b)$ with $1 \leq a < b \leq 3$.
We have then $m \mid f(b) - f(a) = k(124^b - 124^a)$.
If $b - a = 1$, then we have $m \mid 123$. It follows that for any integer $c$, we have $f(c) \equiv 1 + k \cdot 124^c \equiv 1 + k\mod m$. Therefore $f(c) \equiv f(a) \equiv 0\mod m$ for any $c$.
If $b - a = 2$, then $a, b = 1, 3$ and we have $m \mid 124^2 - 1 = 123 \times 125$. The case $m \mid 123$ is already covered above, hence it only remains to look at the case $m \mid 125$, which forces $m = 5$.
In that case, we have $5 \mid 1 + k \cdot 124^c$ for any odd integer $c$, which implies that $5$ does not divide $1 + k \cdot 124^c$ for any even integer $c$.
Thus we must have $n \mid f(2)$ and $n \mid f(4)$. However, this leads to the same situation as above, now with $a, b = 2, 4$. The same argument then tells us that either $n \mid 123$ (which leads to $n \mid f(c)$ for all $c$) or $n \mid 125$ (which leads to $n = 5$, a contradiction with the assumption $m \neq n$).

As can be seen from the proof, one may replace the number $124$ with any number of the form $p^e - 1$ with $p$ prime.
