Lazy oats

All of the $S_p$ are polynomials in $n$ . The first few are $S_0(n)=n,\;\;\;S_1(n)=\frac12 n^2+\frac12 n,\;\;\;S_2(n)=\frac13 n^3+\frac12 n^2+ \frac16 n,\;\;\;S_3(n)=\frac14 n^4+\frac12 n^3+\frac14n$ These are normally written in a factorised form, but in this form we note that the leading term appears to be $\frac{1}{p+1} n^{p+1}$ . This is familiar as the result of integrating the function $n^p$ , which makes sense; it's easy to prove this either by considering the limit as $n \to \infty$ or by using the method of differences .

Now, let's say we have a solution to the problem, $p$ ; ie for some non-negative integers $a_1,a_2,\cdots,a_k$ we have $S_p(n) \equiv S_{a_1} (n)\cdot S_{a_2} (n) \cdots S_{a_k} (n)$

Comparing the leading terms, $\frac{1}{p+1} n^{p+1} \equiv \frac{1}{\left(a_1+1 \right)\left(a_2+1 \right)\cdots \left(a_k+1 \right)} n^{a_1+a_2+\cdots +a_k+k}$

The $1$ s in this expression make it untidy; if we instead write $b_i=a_i+1$ and $q=p+1$ it's just $\frac{1}{q} n^{q} \equiv \frac{1}{b_1 \cdot b_2\cdots b_k} n^{b_1+b_2+\cdots +b_k}$

Now, both the power and the coefficient need to match; so $q=\sum_1^k b_i=\prod_1^k b_i$ and we're looking for sets of numbers whose sum is equal to their product.

In the $p=3$ case, this is $2+2=2\times 2=4$ .

Let's assume $b_1 \le b_2 \le \cdots b_k$ . Let's also assume (for now) that $b_1>1$ .

Now, $\prod_1^k b_i \ge 2^{k-1} b_k$ and $\sum_1^k b_i \le kb_k$

so for any solutions to exist, we need $2^{k-1} b_k \le kb_k$ ie $2^{k-1} \le k$

$2^{k-1}$ grows far more quickly than $k$ ; the only solutions are $k=1$ (not allowed) and $k=2$ . It's easy to check that the only solution with $k=2$ corresponds to the example in the question.

Now, we've only proved that no other solutions exist when $b_1>1$ . There are, in fact, infinitely many sets of positive integers with equal sum and product if you allow $b_1=1$ ; for instance $(1,2,3)$ , $(1,1,2,4)$ and so on.

However, $b_1=1$ implies that $S_0(n)$ divides $S_p(n)$ . But $S_0(n)=1^0+2^0+\cdots +n^0=n$ ; this provides a counterexample, because $S_p(2)$ is odd whenever $p>0$ .

So the only solution is the example given in the question.