Problem 2104

If $a = 1$, the identity of the problem holds regardless of how other positive integers $n, p, q, r$ are chosen. So, we assume that $a \ge 2$ in the sequel. Since the given identity is symmetric in $p, q, r$, we may assume that $p \le q \le r$ holds.

We can rewrite the given identity in the form
$$a^n = a^{p+q+r} - (a^{p+q} + a^{p+r} + a^{q+r}) + (a^p + a^q + a^r).$$
Since $a \ge 2$, we have $a^{p+q} > a^p$, $a^{q+r} > a^q$, $a^{p+r} > a^r$. This together with the fact that $a^p + a^q + a^r > 0$ gives us
$$a^{p+q+r} > a^n > a^{p+q+r} - (a^{p+q} + a^{p+r} + a^{q+r}).$$
From the left half of the inequality above, we see that $n \le p+q+r-1$ and using this we get from the right half of the inequality above that $a^{p+q+r-1} > a^{p+q+r} - (a^{p+q} + a^{p+r} + a^{q+r})$, which simplifies to
$$a^{-1} + a^{-p} + a^{-q} + a^{-r} > 1.$$
Because of the fact that $1 \le p \le q \le r$, we have $a^{-1} \ge a^{-p} \ge a^{-q} \ge a^{-r}$ and since the sum of these 4 numbers is greater than 1 as we saw above, we get $a^{-1} > \frac{1}{4}$. Thus we have $a < 4$ and since $a \ge 2$, we must have either $a = 2$ or $a = 3$.

(1) Consider the case when $a = 3$:

In this case we have $3^{-p} + 3^{-q} + 3^{-r} > 1 - 3^{-1} = \frac{2}{3}$ so that $3^{-p} > \frac{2}{9}$, and therefore, $p = 1$. We then get $3^{-q} + 3^{-r} > \frac{1}{3}$, which in turn implies $3^{-q} > \frac{1}{6}$, and therefore, $q = 1$. Now from $p = q = 1$, we conclude from the given identity that $3^n = 4 \cdot 3^r - 3$ must hold. Since $4 \cdot 3^r < 9 \cdot 3^r = 3^{r+2}$, we must have $n \le r+1$. Consequently, we must have $3^{r+1} \ge 4 \cdot 3^r - 3$, which simplifies to $3^r \le 3$. Therefore, $r = 1$ must hold. Thus we see that if $a = 3$, then $p = q = r = 1$ must hold and then the given identity forces $n$ to satisfy $3^n - 1 = (3-1)^3$, from which we get $n = 2$. Thus we conclude that $(3, 2, 1, 1, 1)$ is the unique solution in the case (1).

(2) Consider the case when $a = 2$:

In this case, we have $2^{-p} + 2^{-q} + 2^{-r} > \frac{1}{2}$, which implies $2^{-p} > \frac{1}{6}$. Therefore, we must have either $p = 1$ or $p = 2$.

* Case when $p = 2$:

In this case, we have $2^{-q} + 2^{-r} > \frac{1}{4}$ so that $2^{-q} > \frac{1}{8}$ and combining with the fact $q \ge p = 2$ we get $q = 2$ as well. Then, we see that the given identity takes the form $2^n = 9 \cdot 2^r - 8$. From $9 \cdot 2^r < 16 \cdot 2^r = 2^{r+4}$ we get $2^{r+3} \ge 9 \cdot 2^r - 8$, which simplifies to $2^r \le 8$. So, we must have $r = 2$ or $r = 3$. Checking each of these cases, we can conclude that $(a, n, p, q, r) = (2, 6, 2, 2, 3)$ is the unique solution in this case.

* Case when $p = 1$:

In this case the given identity takes the form $2^n = 2^{q+r} - 2^q - 2^r + 2$. Since the right side of this equality is less than $2^{q+r}$, we must have $2^{q+r-1} \ge 2^{q+r} - 2^q - 2^r + 2$, which reduces to the inequality $2^{q+r-1} + 2 \le 2^q + 2^r$, from which we get, as $2^q + 2^r \le 2^{r+1}$, that $2^{q+r-1} < 2^{r+1}$. Hence, $q+r-1 < r+1$ and we get $q = 1$. The given identity now reduces, as $p = q = 1$, to the equality $2^n = 2^r$, so we must have $n = r$, which can be any positive integer. Thus the solution in this case takes the form $(a, n, p, q, r) = (2, k, 1, 1, k)$, where $k$ is an arbitrary positive integer.

Summarizing the arguments given above, we can write down all the quintuples $(a, n, p, q, r)$ satisfying the given identity as follows:

* $a = 1$ and $n, p, q, r$ can be arbitrary positive integers.
* $(a, n, p, q, r) = (3, 2, 1, 1, 1)$.
* $(a, n, p, q, r) = (2, 6, 2, 2, 3)$ and its variants obtained by permuting $p, q, r$.
* $(a, n, p, q, r) = (2, k, 1, 1, k)$ where $k$ is an arbitrary positive integer and their variants obtained by permuting $p, q, r$.