International Mathematical Olympiad 2015 Problem 2

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.

Source: International Mathematical Olympiad



We may assume that $a \leq b \leq c$, such 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.

Now let's consider the remaining cases. 


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 $3 \leq a < b \leq c$.


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

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