2015 IMO Problems/Problem 2

Problem

Determine all triples of positive integers $(a,b,c)$ 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 ).

Solution

The solutions for $(a,b,c)$ are $(2,2,2)$, $(2,2,3)$, $(2,6,11)$, $(3,5,7)$, and permutations of these triples.

Throughout the proof, we assume $a \leq b \leq c$, so that $ab-c = 2^m$, $ca-b = 2^n$, $bc-a=2^p$, with $m \leq n \leq p$. Note that $a>1$ since otherwise $b-c=2^m$, which is impossible. Hence $2^n = ac-b \geq (a-1)c \geq 2$, i.e., $n$ and $p$ are positive.

Observe that if $a=b\geq 3$, we get $a(c-1)=2^n$, so $a$ and $c-1$ are (even and) powers of $2$. Hence $c$ is odd and $a^2-c=2^m=1$. Hence $c+1=a^2$ is also a power of $2$, which implies $c=3$. But $a=b=c=3$ is not a solution; hence $a=b\geq 3$ is infeasible. We consider the remaining cases as follows.

Case 1: $a=2$. We have $$2b-c=2^m,\; 2c-b=2^n,\; bc-2=2^p.$$ From the second equation, $b$ is even. From the third equation, if $p=1$, then $b=c=2$; if $p>1$, then $c$ is odd, which implies that $m=0$. Hence $3b=2^n+2$ (so $n\geq 2$), $3c=2^{n+1}+1$, and $(2^{n-1}+1)(2^{n+1}+1)=9(2^{p-1}+1)$. Hence $1\equiv 9 \mod 2^{n-1} \implies n \leq 4$. Hence $n$ is 2 or 4, and $(b,c)$ equals $(2,3)$ or $(6,11)$. Thus the solutions for $(a,b,c)$ are $(2,2,2)$, $(2,2,3)$ or $(2,6,11)$.

Case 2: $3\leq a. Since $(a-1)c \leq 2^n$, we have $c\leq 2^{n-1}$. Hence $$b+a < 2c\leq 2^{n+1}/(a-1) \leq 2^n,$$ $$b-a < c \leq 2^{n-1}.$$ Hence $b-a$ is not divisible by $2^{n-1}$, and $b+a$ is not divisible by $2^{n-1}$ for $a\geq 5$. Adding and subtracting $ac-b=2^n$ and $bc-a=2^p$, we get $$(c-1)(b+a) = 2^p+2^n,$$ $$(c+1)(b-a) = 2^p-2^n.$$ From the latter equation, $c+1$ is divisible by $4$. Hence $c-1$ is not divisible by $4$, which implies that $b+a < 2^n$ is a multiple of $2^{n-1}$. Hence $a \leq 4$ and $b+a=2^{n-1}$.

Consider $a=3$, which implies $3b-c=2^m$, $3c-b=2^n$, $b=2^{n-1}-3$. Hence $2^{n-1}-3=3.2^{m-3}+2^{n-3}$, or $2^{n-3}=2^{m-3}+1$. Hence $m=3$, $n=4$, $b=5$ and $c=7$.

Finally, consider $a=4$, $4c-b=2^n$, $b=2^{n-1}-4$. Hence $c=3.2^{n-3}-1$. But $b \leq c$ implies $2^{n-3} \leq 3$ and $a implies $2^{n-3}>2$. Hence there are no solutions with $a=4$.

We obtain $(a,b,c)=(3,5,7)$ as the only solution with $3 \leq a < b \leq c$.