Problem 1662

1. Initial Setup and Definitions:
- Let $\tau(n)$ denote the number of positive integer divisors of $n$.
- Let $\sigma(n)$ denote the sum of the positive integer divisors of $n$.
- Let $\varphi(n)$ denote the number of positive integers less than or equal to $n$ that are relatively prime to $n$.
- We need to prove that for any integers $a, b > 1$, Brandon can make the calculator display $b$ after a finite sequence of button presses starting from $a$.

2. Reduction to a Simpler Problem:
- Notice that repeatedly applying $\tau$ to any number $n$ will eventually reduce it to $2$. This is because $\tau(n)$ is always at least $2$ for any $n > 1$ and $\tau(2) = 2$.
- Therefore, it suffices to show that from $2$, Brandon can reach any positive integer $b > 1$.

3. Reaching Powers of 2:
- We need to show that Brandon can reach every power of $2$.
- Observe that $\tau(2^k) = k + 1$. Hence, by applying $\tau$ to $2^k$, we can reach $k + 1$.
- Conversely, $\varphi(2^k) = 2^{k-1}$. Thus, by applying $\varphi$ to $2^k$, we can reach $2^{k-1}$.
- By iterating this process, we can reach arbitrarily large powers of $2$.

4. Reaching Arbitrarily Large Numbers:
- To show that we can reach numbers with arbitrarily large $\nu_2$ (the exponent of $2$ in the prime factorization), consider the effect of $\varphi$ on a number $n$.
- Applying $\varphi$ to a number $n$ decreases its $\nu_p$ (exponent of any prime $p$) by at most $1$.
- Repeatedly applying $\varphi$ will eventually reduce $\nu_p$ to $0$ for any prime $p$.

5. Contradiction Argument:
- Suppose for contradiction that the $\nu_p$ of any reachable number is bounded for any prime $p$.
- Consider the number $m$ after repeatedly applying $\varphi$ until it is of the form $2^i3^j$, $2^ip$ where $p \neq 3$ is an odd prime, or $2^i$.
- The $\nu_2$ of $m$ decreases by at most $1$ each time $\varphi$ is applied, and increases by some number $l$.
- If $l = 1$ for two consecutive applications of $\varphi$, then $m$ must have exactly one odd prime divisor $p$ without multiplicity, which is $3 \pmod{4}$ and not equal to $3$.
- However, $\varphi(m)$ would then be divisible by at least two distinct odd primes, contradicting the assumption that $l = 1$ for two consecutive applications.

6. Reaching Arbitrarily Large Numbers:
- Since $\sigma(n)$ generates arbitrarily large numbers, we can reach arbitrarily large numbers.
- For large enough reachable $n$, repeated applications of $\varphi$ starting from $n$ must result in a number of the form $2^ip_n$ for some prime $p_n > 3$.
- This implies that $\frac{n}{f(n)}$ (where $f(n)$ is the largest prime divisor of $n$) is bounded above, leading to a contradiction.

7. Conclusion:
- Therefore, Brandon can reach any positive integer $b > 1$ from any starting integer $a > 1$ using a finite sequence of button presses.

$\blacksquare$