Let's determine the triples of positive integers $(a, b, c)$ for which the numbers

$$
a b-c, \quad b c-a, \quad c a-b
$$

are all powers of 2.

(A power of 2 is an integer of the form $2^{n}$, where $n$ is a non-negative integer.)
Szabó Barnabás's solution. As usual, $v_{2}(x)$ denotes the exponent of 2 in the prime factorization of a positive integer $x$. In the following, we will use the well-known statement without reference, which states that if $v_{2}(x) > v_{2}(y)$, then $v_{2}(x \pm y) = v_{2}(y)$, and if $v_{2}(x) = v_{2}(y) = t$, then $v_{2}(x \pm y) \geq t+1$. If $a=1$, then one of $a b-c$ and $a c-b$ is not positive, so it cannot be a power of 2. Therefore, $a \neq 1$, similarly $b, c \neq 1$.

1. Case: all variables are even. By symmetry, we can assume that $v_{2}(a) \geq v_{2}(b) \geq v_{2}(c) \geq 1$. Then $v_{2}(a b-c) = v_{2}(c)$, but $a b-c$ is a power of 2, so $a b-c \leq c$, i.e., $a b \leq 2 c$. From $v_{2}(a c-b) = v_{2}(b)$, we get $a c \leq 2 b$. Multiplying the two inequalities, we get $a \leq 2$, but $a \neq 1$, so $a=2$. Substituting this, we get $2 b \leq 2 c$ and $2 c \leq 2 b$, i.e., $b=c$, but then $v_{2}(b c-a) = v_{2}(b^2 - 2) = 1$, so $b^2 - 2 = 2^1$, from which $b=c=2$. The first case thus gives the triplet $(a, b, c) = (2,2,2)$, which is indeed a valid solution.
2. Case: one variable is odd. We can assume that $c$ is odd. Suppose $v_{2}(a) \neq v_{2}(b)$, say $v_{2}(a) > v_{2}(b)$. Then $a$ is even, so $a b-c$ is odd, i.e., $a b-c = 1$. On the other hand, $v_{2}(b c-a) = v_{2}(b)$, so $b c-a = 2^{v_{2}(b)}$ is a divisor of $b$, $b c-a \leq b$. Then $a b-1 = c \leq \frac{b+a}{b} \leq 1+a$, from which $a(b-1) \leq 2$, so $a=2$ and $b=2$, which contradicts $v_{2}(a) > v_{2}(b)$.

Therefore, $t = v_{2}(a) = v_{2}(b)$. First, suppose that $t \geq 1$. Then $2 \mid a b$, so $a b-c = 1$. Substituting $c$,

\[
\begin{aligned}
& b c-a = a b^2 - (a+b) = 2^x \\
& c a-b = a^2 b - (a+b) = 2^y
\end{aligned}
\]

where we can assume $x \leq y$. Subtracting the two, we get $2^x (2^{y-x} - 1) = a b (a - b)$. If $x = y$, then $a = b$ and $a^3 - 2a = a (a^2 - 2) = 2^x$. Since $a$ is even, $v_{2}(a^2 - 2) = 1$, so $a^2 - 2 = 2^1$, giving $a = 2$. This yields the triplet $(3,2,2)$.

If $x < y$, then $v_{2}(a - b) = v_{2}(2a - (a+b)) = t+1$, so $x = v_{2}(a b (a - b)) = 3t + 1$. Let $a b^2 = 2^{3t} d$ and $a + b = 2^{3t} e$, then substituting and dividing by $2^{3t}$, we get $d - e = 2$, so $\frac{d}{e} \leq 3$, hence $a b^2 \leq 3(a + b)$. From $b \geq 2$, we get $a \leq \frac{3}{b^2} a + \frac{3}{b} \leq \frac{3}{4} a + \frac{3}{2}$, so $a \leq 6$. If $a = 2$, then $2 b^2 \leq 6 + 3b$, so $b = 2$ (since $v_{2}(b) = 1$), yielding the triplet $(3,2,2)$ again. Similarly, if $a = 4$, there is no solution, while if $a = 6$, we get the triplet $(2,6,11)$.

Now we consider the case where $t = 0$, i.e., all three numbers are odd. Let $b c - a = 2^x$, $c a - b = 2^y$, $a b - c = 2^z$. We can assume $x \leq y \leq z$. Then $2^y \mid (a b - c) - (a c - b) = (b - c)(a + 1)$ and $2^y \mid (a b - c) + (a c - b) = (b + c)(a - 1)$. Clearly, $v_{2}(b - c) = 1$ or $v_{2}(b + c) = 1$, so $a \geq 2^{y-1} - 1$. But $(a + b)(c - 1) = 2^x + 2^y \leq 2^{y+1}$, and $a + b \geq 2^{y-1}$, so $c \leq 5$. If $c = 5$, then $b = 1$, which is a contradiction. Therefore, only $c = 3$ is possible. Since $b > 1$, $a < 2^y - 1$, so $a = 2^{y-1} + 1$ or $a = 2^{y-1} - 1$. From the former, we get a contradiction after a simple calculation, while from the latter, we get the solution $(3,5,7)$, which is indeed valid. The solutions are thus: $(2,2,2)$, $(3,2,2)$, $(2,6,11)$, and $(3,5,7)$, and of course their permutations.