Finding $(a, b, c)$ with $ab-c$, $bc-a$, and $ca-b$ being powers of $2$ This is a 2015 IMO problem.  It seems difficult to solve.

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

Four such triples are $(a,b,c)=(2,2,2),(2,3,2),(3,5,7),(2,6,11)$.
 A: A Sketch of a Possible Solution
Wlog order the solutions so that $a \ge b \ge c \ge 1$ with the system of equations
$\begin{align}
ab - c &= 2^k \\
ac - b &= 2^l \\
bc - a &= 2^m
\end{align}$
with $k \ge l \ge m \ge 0$. 
We distinguish four cases:


*

*$a=b=c$

*$a=b>c$

*$a>b=c$

*$a>b>c$


Case 1: $a=b=c$
The main system of equations reduce to the single equation
$a(a-1) = 2^k \tag{1A}$. 
$a,a-1$ cannot both be powers of 2 unless $a=2$. Therefore, the only solution for this case is 
$\boxed{a=b=c=2}$
Case 2: $a=b>c$
Main equations reduce to
$\begin{align}
a^2 - c &= 2^k \tag{2A}\\
a(c - 1) &= 2^l \tag{2B}
\end{align}$
with $a > c, k > l$. Equation (2B) implies that $a$ is even since $a>c$, and (2A) now implies that $c$ is even since $k>l\ge0$. Then $a=2^l$ and so (2A) becomes: $2^{2l} = 2^k+2$ which can only be satisfied by $k=l=1$. However, this violates $k>l$ (it is just the first case again). No solutions for this case.
Case 3: $a>b=c$
Main equations reduce to
$\begin{align}
b(a - 1) &= 2^k \tag{3A}\\
b^2 - a &= 2^m \tag{3B}
\end{align}$
with $a > b, k > m$. Since $a>1$, (3A) implies $a$ is odd. On the other hand, (3B) implies $b>1$, so by (3A) $b$ must be even and $m=0$. Furthermore, we can put $a=2^p+1,b=2^q$ with $p\ge1,q\ge1$. So by (3B)
$b^2-a = 2^{2q}-(2^p+1) = 1$ so $2^{2q}-2^p = 2$. Whence $p=q=1$ must hold. So $a=3,b=2$ and the only solutions for this case are
$\boxed{a=3,b=c=2}$
Case 4: $a>b>c$
We restate the system of equations
$\begin{align}
ab - c &= 2^k \\
ac - b &= 2^l \tag{4A}\\
bc - a &= 2^m
\end{align}$
where we now have $a>b>c\ge1,k>l>m\ge0$. If exactly one of $a,b,c$ is even, then $ab-c,ac-b,bc-a$ are all odd, which cannot satisfy the system of equations with $k>l>m$ (see Theo Bendit's argument).
So now distinguish three other cases:


*

*$a,b,c$ all even

*Two of $a,b,c$ are even

*$a,b,c$ all odd  


Case 4.1: $a>b>c$ with $a,b,c$ all even
We write $a=2p,b=2q,c=2r$ with $p>q>r\ge1$. Then system (4A) can be re-expressed as
$\begin{align}
4pq - 2r &= 2^k \\
4pr - 2q &= 2^l \\
4qr - 2p &= 2^m
\end{align}$
Since the left-hand sides are all integers, divide through by 2 to get:
$\begin{align}
2pq - r &= 2^{k-1} \\
2pr - q &= 2^{l-1} \\
2qr - p &= 2^{m-1}
\end{align}$
Since $p>q>r$, only $r$ can be odd. But if this is so, $m=1$ and this is excluded since $2pq-r\ge2\cdot3\cdot2-1=11$. So $p,q,r$ are all even. Then write $p=2s,q=2t,r=2u$ with $s>t>u$. Then system (4A) becomes
$\begin{align}
4pq - r &= 2^{k-2} \\
4pr - q &= 2^{l-2} \\
4qr - p &= 2^{m-2}
\end{align}$
Using similar arguments we can show that $s,t,u$ must all be even. Since these arguments can be repeated ad infinitum, we conclude that there are no solutions for Case 4.1.
Case 4.2: $a>b>c$ with two of $a,b,c$ even
In this case, exactly one of $ab-c,ac-b,bc-a$ are odd, so to satisfy (4A) it must be the smallest of these: $bc-a$. Hence we must have $a$ odd, and $b,c$ even. Furthermore, $m=0$. So in (4A) we must have
$bc - a = 1 \tag{4.2A}$ with the restriction that $b>c\ge2$.
Now substitute the equivalent expression for $a$ into (4A) to get
$\begin{align}
b^2c - b - c &= 2^k \\
bc^2 - b -c &= 2^l \tag{4.2B}\\
\end{align}$
Hence by subtraction
$\begin{align}
b^2c - bc^2 &= 2^k - 2^l \\
bc(b-c) &= 2^l(2^{k-l}-1) \tag{4.2C}\\
\end{align}$
This is satisfied by $b=6,c=2$ or the solution $(a,b,c)=(11,6,2)$.
There may be other solutions.

A: Without loss of generality: replace $b$ with $a+k1$ and $c$ with $a+ k2$
$(a, a+k1, a +k2) $
$ab- c = a2 + k1a -a -k2 $
$bc- a = a2 + k1k2 + (k1+k2 -1)a$
$ca- b = a2+ (k2-1)a - k1$
All these must by $2N$ for some $n$ in $I > 0$
