A number theory problem by Priyanshu Mishra

First suppose $m > n.$ Let $a=m-n.$ The equation becomes $2^a2^n n - 2^n(a+n) = 2^k.$ Now $n>k$ is impossible because the left side is divisible by $2^n,$ so the right side must be as well. Let $d = k-n \ge 0.$ Then divide both sides by $2^n$ to get $n(2^a-1)-a = 2^d.$ Look at this equation modulo $2^a-1$ : we get $2^d \equiv -a,$ but the powers of $2$ modulo $2^a-1$ are $1,2,\ldots,2^{a-1}.$ One of these must be congruent to $-a:$ $2^b+a \equiv 0 \pmod{2^a-1}$ for some $b \le a-1.$ In particular, $2^{a-1}+a$ must be at least $2^a-1.$ So $a \ge 2^{a-1}-1.$ This holds only for $a=1,2,3.$

If $a=1$ we get $n-1$ is a power of $2,$ hence the family of solutions $(k,m,n) = (2,3,2),(4,4,3),(7,6,5),(12,10,9),\ldots,(71,66,65).$ There are seven of these.

If $a=2$ we get $3n-2$ is a power of $2,$ and it's not hard to write down the powers of $2$ that are congruent to $1$ mod $3$ and solve for $n$ : the solutions are $n=1,2,6,22,86,$ leading to $(1,3,1),(4,4,2),(10,8,6),(28,24,22),(94,88,86).$ There are five of these.

If $a=3$ we get $7n-3$ is a power of $2.$ The powers of $2$ congruent to $4$ mod $7$ are precisely $2^{3c+2},$ with corresponding values of $n$ given by $n=1,5,37,$ leading to $(3,4,1), (10,8,5), (45,40,37).$ There are three of these.

Now suppose $m < n$ ( $m=n$ is impossible). Let $x=n-m.$ The equation becomes $2^m(x+m)-2^x 2^m m = 2^k.$ Again $m \le k$ is mandatory, so divide out by $2^m$ to get $x+(1-2^x)m = 2^e$ where $e = k-m.$ Now $x + (1-2^x)m \le x + 1-2^x \le 0$ for all positive $x$ if $m \ge 1,$ so the only possibility is $m=0.$ This gives the family of solutions where $x=n=$ a power of $2$ : $(0,0,1),(1,0,2),(2,0,4),\ldots,(6,0,64).$ There are seven of these.

The final answer is $7+5+3+7=\fbox{22}.$