To prove that the set $S$ of all pairs of positive integers $(m, n)$ such that $m \neq n$ and $a_{m+n} = a_m + a_n$ is infinite, we'll construct an infinite subset of $S$ by considering pairs of the form $(p^k, p^l)$, where $p$ is a prime number and $k, l$ are distinct positive integers.

Step 1: Calculate $a_m$, $a_n$, and $a_{m+n}$

Consider $m = p^k$ and $n = p^l$, where $p$ is a prime and $k$ and $l$ are distinct positive integers. For any positive integer power of a prime number, the number of distinct prime divisors is always 1. Therefore: $a_m = a_{p^k} = 1$ $a_n = a_{p^l} = 1$

Now, let's calculate $a_{m+n} = a_{p^k + p^l}$.

Step 2: Analyze $m+n = p^k + p^l$

Without loss of generality, assume $k < l$. We can factor $p^k + p^l$ as: $p^k + p^l = p^k(1 + p^{l-k})$ Here, $p^k$ contributes exactly one prime factor, which is $p$. The term $1 + p^{l-k}$ is an integer greater than 1. Since $l-k$ is a positive integer, $1 + p^{l-k}$ is not divisible by $p$. Therefore, $1 + p^{l-k}$ and $p^k$ are coprime.

Since $1 + p^{l-k}$ is not divisible by $p$ and does not introduce any new prime divisors, $p^k + p^l$ has exactly two distinct prime divisors: $p$ from the $p^k$ term and whatever prime divides $1 + p^{l-k}$ (note that $1 + p^{l-k}$ is not divisible by $p$, and thus must be divisible by some other prime).

Therefore: $a_{m+n} = a_{p^k + p^l} = 2$

Step 3: Check the condition $a_{m+n} = a_m + a_n$

We already know that: $a_m = 1, \quad a_n = 1$ $a_{m+n} = 2$ Thus: $a_{m+n} = 2 = 1 + 1 = a_m + a_n$ So, $(m, n) = (p^k, p^l)$ satisfies the condition $a_{m+n} = a_m + a_n$.

Step 4: Infinite Set $S$

Since there are infinitely many primes $p$, and for each prime $p$, we can choose infinitely many distinct pairs of positive integers $k$ and $l$ with $k \neq l$, we can generate infinitely many pairs $(m, n) = (p^k, p^l)$ such that $a_{m+n} = a_m + a_n$.

Therefore, the set $S$ is infinite.

Conclusion

We have shown that $S$, the set of all pairs of positive integers $(m, n)$ such that $m \neq n$ and $a_{m+n} = a_m + a_n$, is infinite.

Do you have any questions, or would you like further details?

Here are some related questions for further exploration:

1. What happens if $m$ and $n$ are not powers of the same prime?
2. How does the condition $a_{m+n} = a_m + a_n$ change for composite numbers?
3. What is the smallest pair $(m, n)$ that belongs to the set $S$?
4. Can we generalize this problem for higher powers