Subfield: Number Theory


Question: Determine all triples $(a, b, c)$ of positive integers for which $a b-c, b c-a$, and $c a-b$ are powers of 2 .

Explanation: A power of 2 is an integer of the form $2^{n}$, where $n$ denotes some nonnegative integer.

Solution:
It can easily be verified that these sixteen triples are as required. Now let $(a, b, c)$ be any triple with the desired property. If we would have $a=1$, then both $b-c$ and $c-b$ were powers of 2 , which is impossible since their sum is zero; because of symmetry, this argument shows $a, b, c \geqslant 2$.

Case 1. Among $a, b$, and $c$ there are at least two equal numbers.

Without loss of generality we may suppose that $a=b$. Then $a^{2}-c$ and $a(c-1)$ are powers of 2. The latter tells us that actually $a$ and $c-1$ are powers of 2 . So there are nonnegative integers $\alpha$ and $\gamma$ with $a=2^{\alpha}$ and $c=2^{\gamma}+1$. Since $a^{2}-c=2^{2 \alpha}-2^{\gamma}-1$ is a power of 2 and thus incongruent to -1 modulo 4 , we must have $\gamma \leqslant 1$. Moreover, each of the terms $2^{2 \alpha}-2$ and $2^{2 \alpha}-3$ can only be a power of 2 if $\alpha=1$. It follows that the triple $(a, b, c)$ is either $(2,2,2)$ or $(2,2,3)$.

Case 2. The numbers $a, b$, and $c$ are distinct.

Due to symmetry we may suppose that

$$
2 \leqslant a<b<c .
\tag{1}
$$

We are to prove that the triple $(a, b, c)$ is either $(2,6,11)$ or $(3,5,7)$. By our hypothesis, there exist three nonnegative integers $\alpha, \beta$, and $\gamma$ such that

$$
b c-a  =2^{\alpha}, \tag{2}
$$
$$
a c-b  =2^{\beta}, \tag{3}
$$
$$
\text { and } \quad a b-c =2^{\gamma} .\tag{4}
$$

Evidently we have

$$
\alpha>\beta>\gamma\tag{5}
$$

Depending on how large $a$ is, we divide the argument into two further cases.

Case 2.1. $\quad a=2$.

We first prove that $\gamma=0$. Assume for the sake of contradiction that $\gamma>0$. Then $c$ is even by (4) and, similarly, $b$ is even by (5) and (3). So the left-hand side of (2) is congruent to 2 modulo 4 , which is only possible if $b c=4$. As this contradicts (1), we have thereby shown that $\gamma=0$, i.e., that $c=2 b-1$.

Now (3) yields $3 b-2=2^{\beta}$. Due to $b>2$ this is only possible if $\beta \geqslant 4$. If $\beta=4$, then we get $b=6$ and $c=2 \cdot 6-1=11$, which is a solution. It remains to deal with the case $\beta \geqslant 5$. Now (2) implies

$$
9 \cdot 2^{\alpha}=9 b(2 b-1)-18=(3 b-2)(6 b+1)-16=2^{\beta}\left(2^{\beta+1}+5\right)-16,
$$

and by $\beta \geqslant 5$ the right-hand side is not divisible by 32 . Thus $\alpha \leqslant 4$ and we get a contradiction to (5).



Case 2.2. $a \geqslant 3$.

Pick an integer $\vartheta \in\{-1,+1\}$ such that $c-\vartheta$ is not divisible by 4 . Now

$$
2^{\alpha}+\vartheta \cdot 2^{\beta}=\left(b c-a \vartheta^{2}\right)+\vartheta(c a-b)=(b+a \vartheta)(c-\vartheta)
$$

is divisible by $2^{\beta}$ and, consequently, $b+a \vartheta$ is divisible by $2^{\beta-1}$. On the other hand, $2^{\beta}=a c-b>$ $(a-1) c \geqslant 2 c$ implies in view of (1) that $a$ and $b$ are smaller than $2^{\beta-1}$. All this is only possible if $\vartheta=1$ and $a+b=2^{\beta-1}$. Now (3) yields

$$
a c-b=2(a+b),
\tag{6}
$$

whence $4 b>a+3 b=a(c-1) \geqslant a b$, which in turn yields $a=3$.

So (6) simplifies to $c=b+2$ and (2) tells us that $b(b+2)-3=(b-1)(b+3)$ is a power of 2. Consequently, the factors $b-1$ and $b+3$ are powers of 2 themselves. Since their difference is 4 , this is only possible if $b=5$ and thus $c=7$. Thereby the solution is complete.
As in the beginning of the first solution, we observe that $a, b, c \geqslant 2$. Depending on the parities of $a, b$, and $c$ we distinguish three cases.

Case 1. The numbers $a, b$, and $c$ are even.

Let $2^{A}, 2^{B}$, and $2^{C}$ be the largest powers of 2 dividing $a, b$, and $c$ respectively. We may assume without loss of generality that $1 \leqslant A \leqslant B \leqslant C$. Now $2^{B}$ is the highest power of 2 dividing $a c-b$, whence $a c-b=2^{B} \leqslant b$. Similarly, we deduce $b c-a=2^{A} \leqslant a$. Adding both estimates we get $(a+b) c \leqslant 2(a+b)$, whence $c \leqslant 2$. So $c=2$ and thus $A=B=C=1$; moreover, we must have had equality throughout, i.e., $a=2^{A}=2$ and $b=2^{B}=2$. We have thereby found the solution $(a, b, c)=(2,2,2)$.

Case 2. The numbers $a, b$, and $c$ are odd.

If any two of these numbers are equal, say $a=b$, then $a c-b=a(c-1)$ has a nontrivial odd divisor and cannot be a power of 2 . Hence $a, b$, and $c$ are distinct. So we may assume without loss of generality that $a<b<c$.

Let $\alpha$ and $\beta$ denote the nonnegative integers for which $b c-a=2^{\alpha}$ and $a c-b=2^{\beta}$ hold. Clearly, we have $\alpha>\beta$, and thus $2^{\beta}$ divides

$$
a \cdot 2^{\alpha}-b \cdot 2^{\beta}=a(b c-a)-b(a c-b)=b^{2}-a^{2}=(b+a)(b-a) .
$$

Since $a$ is odd, it is not possible that both factors $b+a$ and $b-a$ are divisible by 4 . Consequently, one of them has to be a multiple of $2^{\beta-1}$. Hence one of the numbers $2(b+a)$ and $2(b-a)$ is divisible by $2^{\beta}$ and in either case we have

$$
a c-b=2^{\beta} \leqslant 2(a+b) .
\tag{7}
$$

This in turn yields $(a-1) b<a c-b<4 b$ and thus $a=3$ (recall that $a$ is odd and larger than 1). Substituting this back into (7) we learn $c \leqslant b+2$. But due to the parity $b<c$ entails that $b+2 \leqslant c$ holds as well. So we get $c=b+2$ and from $b c-a=(b-1)(b+3)$ being a power of 2 it follows that $b=5$ and $c=7$.

Case 3. Among $a, b$, and $c$ both parities occur.

Without loss of generality, we suppose that $c$ is odd and that $a \leqslant b$. We are to show that $(a, b, c)$ is either $(2,2,3)$ or $(2,6,11)$. As at least one of $a$ and $b$ is even, the expression $a b-c$ is odd; since it is also a power of 2 , we obtain

$$
a b-c=1 \text {. }
\tag{8}
$$

If $a=b$, then $c=a^{2}-1$, and from $a c-b=a\left(a^{2}-2\right)$ being a power of 2 it follows that both $a$ and $a^{2}-2$ are powers of 2 , whence $a=2$. This gives rise to the solution $(2,2,3)$.



We may suppose $a<b$ from now on. As usual, we let $\alpha>\beta$ denote the integers satisfying

$$
2^{\alpha}=b c-a \quad \text { and } \quad 2^{\beta}=a c-b
\tag{9}
$$

If $\beta=0$ it would follow that $a c-b=a b-c=1$ and hence that $b=c=1$, which is absurd. So $\beta$ and $\alpha$ are positive and consequently $a$ and $b$ are even. Substituting $c=a b-1$ into (9) we obtain

$$
2^{\alpha}  =a b^{2}-(a+b), \tag{10}
$$
$$
\text { and } \quad 2^{\beta}  =a^{2} b-(a+b) .
\tag{11}
$$

The addition of both equation yields $2^{\alpha}+2^{\beta}=(a b-2)(a+b)$. Now $a b-2$ is even but not divisible by 4 , so the highest power of 2 dividing $a+b$ is $2^{\beta-1}$. For this reason, the equations (10) and (11) show that the highest powers of 2 dividing either of the numbers $a b^{2}$ and $a^{2} b$ is likewise $2^{\beta-1}$. Thus there is an integer $\tau \geqslant 1$ together with odd integers $A, B$, and $C$ such that $a=2^{\tau} A, b=2^{\tau} B, a+b=2^{3 \tau} C$, and $\beta=1+3 \tau$.

Notice that $A+B=2^{2 \tau} C \geqslant 4 C$. Moreover, (11) entails $A^{2} B-C=2$. Thus $8=$ $4 A^{2} B-4 C \geqslant 4 A^{2} B-A-B \geqslant A^{2}(3 B-1)$. Since $A$ and $B$ are odd with $A<B$, this is only possible if $A=1$ and $B=3$. Finally, one may conclude $C=1, \tau=1, a=2, b=6$, and $c=11$. We have thereby found the triple $(2,6,11)$. This completes the discussion of the third case, and hence the solution. There are sixteen such triples, namely $(2,2,2)$, the three permutations of $(2,2,3)$, and the six permutations of each of $(2,6,11)$ and $(3,5,7)$


Final answer: $(2,2,2),(2,2,3),(2,3,2),(3,2,2),(2,6,11),(2,11,6),(6,2,11),(6,11,2),(11,2,6),(11,6,2),(3,5,7),(3,7,5),(5,3,7),(5,7,3),(7,3,5),(7,5,3)$