Problem 1809

We will prove that there are only two families of solutions:
$$(1, 1, 2^x - 1), \quad (1, 2^x - 1, 2^x + 1),$$
where $x$ is a positive integer.

*Step I:* $a, b, c$ are odd and pairwise coprime in pairs.
It is easy to see that
$$a \equiv b \equiv c \pmod{2}.$$
If $a, b, c$ are all even, denote by $d, e, f$ the exponents of 2 in their prime factorization, respectively, with $1 \le d \le e \le f$. Then the exponent of 2 in $bc+a$ is $d$. Since $bc+a$ is a power of 2, we get
$$bc + a = 2^d,$$
which is impossible because $2^d \le a$. Thus, $a, b, c$ must be odd numbers.
If $a, b$ have a common odd prime factor $p$, then $p$ divides $ac + b$, so $ac + b$ is divisible by $p$ and thus cannot be a power of 2. Similarly, one shows $(a, c) = 1$ and $(b, c) = 1$, which completes Step I.

Hence, we can assume $a \le b \le c$, where $a, b, c$ are odd positive integers pairwise coprime such that
$$ab + c, \quad ac + b, \quad bc + a$$
are powers of 2. Since $a \le b \le c$, we have
$$ab + c \le ac + b \le bc + a.$$
*Step II:* $a = 1$. Suppose, by contradiction, that $1 < a$. Then $1 < a < b < c$, and
$$ab + c = 2^k, \quad ac + b = 2^m, \quad bc + a = 2^n,$$
with $3 \le k < m < n$ (since $2^k \ge 3 + 5 = 8$).
From the above equalities, we get the congruences:
$$ab \equiv -c \pmod{2^k}, \quad ac \equiv -b \pmod{2^k}, \quad bc \equiv -a \pmod{2^k}.$$
Multiplying these congruences and noting that $a, b, c$ are odd, we obtain
$$abc \equiv -1 \pmod{2^k}$$
and
$$a^2 \equiv b^2 \equiv c^2 \equiv 1 \pmod{2^k}$$
(by multiplying the first congruence by $c$, the second by $b$, and the third by $a$).

Since for any odd natural number $x$,
$$\gcd(x - 1, x + 1) = 2,$$
we deduce that
$$a \equiv \pm 1 \pmod{2^{k-1}}, \quad b \equiv \pm 1 \pmod{2^{k-1}}, \quad c \equiv \pm 1 \pmod{2^{k-1}}.$$
From these congruences, it follows that
$$a, b, c \ge 2^{k-1} - 1.$$
This inequality implies
$$2^k = ab + c \ge (2^{k-1} - 1)^2 + 2^{k-1} - 1 = 2^{2k-2} - 2^{k-1},$$
which means $2 \ge 2^{k-1} - 1 \ge 3$, a contradiction since $k > 3$. Therefore, the assumption is false and $a = 1$.

Step III: the solutions. If $b = 1$, then clearly $c = 2^x - 1$. Otherwise, assume $a = 1$ and $1 < b < c$. From Step II, we know that $c > b \ge 2^{k-1} - 1$. Since $b + c = 2^k$ and $b, c$ are odd, the only possibility is $b = 2^{k-1} - 1$, $c = 2^{k-1} + 1$.
Hence, the only solutions are: $(1, 1, 2^x - 1)$ and $(1, 2^x - 1, 2^x + 1)$, $x \in \mathbb{N}^*$.