Problem 1078

If we carry out the indicated product in the following expression:

$$\left\{\begin{array}{l} \left(1-p_{1}+p_{1}^{2}-p_{1}^{3}-p_{1}^{3}+\ldots+(-1)^{\alpha_{1}} p_{1}^{\alpha_{1}}\right) \times \\ \left(1-p_{2}+p_{2}^{2}-p_{2}^{3}-p_{2}^{3}+\ldots+(-1)^{\alpha_{2}} p_{2}^{\alpha_{2}}\right) \times \\ \cdots \\ \left(1-p_{r}+p_{r}^{2}-p_{r}^{3}-p_{r}^{3}+\ldots+(-1)^{\alpha_{r}} p_{r}^{\alpha_{r}}\right) \end{array}\right.$$

we obtain an algebraic sum whose terms will be products formed from the prime divisors $\left(p_{1}, p_{2}, \ldots, p_{3}\right)$, where $p_{1}$ will appear in the $0,1,2, \ldots$ or $\alpha_{1}$-th power; $p_{2}$ in the $0,1,2, \ldots$ or $\alpha_{2}$-th power; ... and finally $p_{r}$ in the $0,1,2, \ldots$ or $\alpha_{r}$-th power. Each term will thus be a divisor of $m$, and since all such products that can be formed under the mentioned conditions are present in the expansion of (1), each divisor of $m$ will appear exactly once among the terms; the divisors will now appear in this sum with a positive sign for which $\alpha_{1}+\alpha_{2}+\ldots$ is even, so $S(d)=1$, while with a negative sign those terms will appear in the expansion for which $S(d)=-1$. Thus, to obtain the desired $\sum S(d)$, it is sufficient to substitute 1 for all $p$'s in (1). In this case, in the expansion of (1), 1 or -1 will appear in place of each term, depending on whether $S(d)=1$ or $S(d)=-1$. Thus, indeed, the expansion of (1) gives the value of $\sum S(d)$ when 1 is substituted for all $p$'s.

The number $m=p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \ldots p_{r}^{\alpha_{r}}$ is a perfect square if and only if all $\alpha$'s are even; in this case, the expressions in each row of (1) will be equal to 1, and thus their product: $\sum S(d)=1$. However, if even one $\alpha$, for example, $\alpha_{r}$, is odd, i.e., $m$ is not a perfect square, then the expression in the $r$-th row of (1) will be 0, and thus the entire expression $\sum S(d)=0$.

(Dénes König, Budapest.)

The problem was also solved by: Bartók I., Dömény I., Haar A., Szücs A.