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: $3\leq a<b\leq c$ . 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 $a<b$ 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 $3 \leq a < b \leq c$ .

2015 IMO (Problems) • Resources
Preceded by Problem 1	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 3
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2015 IMO Problems/Problem 2

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