Finding $(a, b, c)$ with $ab-c$, $bc-a$, and $ca-b$ being powers of $2$
Solution 1:
Suppose $a,b,c\in\mathbb{N}$ and $\alpha,\beta,\gamma\in\mathbb{N}_0$ are such that $bc-a=2^\alpha$, $ca-b=2^\beta$, and $ab-c=2^\gamma$. It is easily seen that either all of $a,b,c$ are odd, or at least two of them are even. We claim that all solutions $(a,b,c)$ are permutations of $(2,2,2)$, $(2,2,3)$, $(2,6,11)$, and $(3,5,7)$.
If two of $a$, $b$, and $c$ are equal, say $a=b$, then we have $2^\alpha=bc-a=a(c-1)$, or $c-1=2^\nu$ and $a=2^{\alpha-\nu}$ for some $\nu\in\mathbb{N}_0$ such that $\nu\leq\alpha$. Hence, $2^{2\alpha-2\nu}=a^2=ab=2^\gamma+c=2^\gamma+2^\nu+1$, which means $a=2$ and $\{\gamma,\nu\}=\{0,1\}$. If $\nu=0$, then $(a,b,c)=(2,2,2)$; if $\nu=1$, then $(a,b,c)=(2,2,3)$. From now on, we assume that $a$, $b$, and $c$ are pairwise distinct. Without loss of generality, assume that $a<b<c$. It follows immediately that $\alpha>\beta>\gamma$.
If $\gamma=0$, we have $ab-c=2^\gamma=1$, or $c=ab-1$. Plugging this in, we have $$2^\alpha=bc-a=b(ab-1)-a=ab^2-(a+b)$$ and $$2^\beta=ca-b=(ab-1)a-b=a^2b-(a+b)\,.$$ If both $a$ and $b$ are odd, then $c$ is even, and we have a contradiction. If $a$ or $b$ is even, then both are even. Suppose that $k$ and $l$ are the largest positive integers such that $2^k\mid a$ and $2^l\mid b$. We claim that $k=l=1$. If $k\neq l$, then from $2^\alpha-2^\beta=ab(a-b)$, we obtain $\beta=k+l+\min\{k,l\}$. Nonetheless, from $2^\alpha+2^\beta=(a+b)(ab-2)$, we get $\beta=1+\min\{k,l\}<k+l+\min\{k,l\}$, which is a contradiction. Thus, $k=l$. Write $a=2^ks$ and $b=2^kt$, where $s$ and $t$ are odd integers. We see that $$2^\alpha+2^\beta=(a+b)(ab-2)=2^{k+1}(s+t)\left(2^{2k-1}st-1\right)$$ and $$2^\alpha-2^\beta=2^{3k+1}st\left(\frac{s-t}{2}\right)\,.$$ This means $\beta\geq 2^{3k+1}$. That is, $2^{2k}\mid s+t$, whence $4\mid s+t$. This would then mean that $\frac{s-t}{2}$ is odd. Hence, $\beta=3k+1$. Now, we have $$2^{3k+1}=2^\beta=a^2b-(a+b)=2^{3k}s^2t-2^k(s+t)\geq2^k\left(2^{2k}-2\right)s^2t\,.$$ If $k>1$, then $s^2t\leq 2$, implying that $s=t=1$, but this violates the assumption that $a<b$. Consequently, $k=1$, so $s^2t\leq 4$, which gives $s=1$ and $t=3$. Thence, $(a,b,c)=(2,6,11)$.
From now on, we assume that $\gamma>0$. If $a$ is even, then $$2^\beta-2^\gamma=(ca-b)-(ab-c)=(c-b)(a+1)$$ leads to $2^\gamma\mid c-b$, and $$2^\beta+2^\gamma=(ca-b)+(ab-c)=(c+b)(a-1)$$ leads to $2^\gamma \mid c+b$. Ergo, $2^\gamma\mid (c+b)-(c-b)=2b$ and $2^\gamma\mid (c+b)+(c-b)=2c$, or $2^{\gamma-1}\mid b$ and $2^{\gamma-1}\mid c$. Now, as $2^\gamma=ab+c$, $2\mid a$, and $2^{\gamma-1}\mid b$, we conclude that $2^\gamma\mid c$. Therefore, $2^\gamma\mid b$ as well, whence $c=2^\gamma w$ for some odd $w\in\mathbb{N}$. Similarly, $2^\beta\mid b$ and $2^\beta\mid a$ (by considering $2^\alpha+2^\beta$ and $2^\alpha-2^\beta$), with $b=2^\beta v$ and $a=2^\delta u$ for some odd $v,u\in\mathbb{N}$ and for some $\delta\in\mathbb{N}$ such that $\delta>\beta$. If $b$ or $c$ is even, we can use a similar method to achieve the same result.
Consequently, in the case where $\gamma>0$, if any of $a$, $b$, and $c$ is even, then $a=2^\delta u$, $b=2^\beta v$, and $c=2^\gamma w$ for some odd $u,v,w\in\mathbb{N}$ and $\delta \in \mathbb{N}$ with $\delta>\beta$. From $2^\beta=ca-b$ and $2^\gamma=ab-c$, we get $$2^\beta=\left(ab-2^\gamma\right)a-b=a^2b-2^\gamma a-b=2^{2\delta+\beta}u^2v-2^{\gamma+\delta} u - 2^\beta v\geq \left(2^{2\delta+\beta}-2^{\gamma+\delta}-2^\beta\right)u^2v\,.$$ Therefore, $2\leq 2^{2\delta-1}u^2v\leq \left(2^{2\delta}-2^{\gamma+\delta-\beta}-1\right)u^2v \leq 1$, which is absurd. That is, $a$, $b$, and $c$ must all be odd.
Now, $\gamma>0$ and all $a$, $b$, and $c$ are odd. From the equality $$2^{2\beta}-2^{2\gamma}=\left(2^\beta-2^\gamma\right)\left(2^\beta+2^\gamma\right)=\big((c-b)(a+1)\big)\big((c+b)(a-1)\big)=\left(c^2-b^2\right)\left(a^2-1\right)\,,$$ we get $2^6\mid 2^{2\beta}-2^{2\gamma}$, making $\gamma \geq 3$. Plugging $c=ab-2^\gamma$ into $bc-a=2^\alpha$ and $ca-b=2^\beta$ to get $\left(\frac{b^2-1}{2^\gamma}\right)a-b=2^{\alpha-\gamma}$ and $\left(\frac{a^2-1}{2^\gamma}\right)b-a=2^{\beta-\gamma}$. Since $\alpha>\beta>\gamma$, we deduce that $\frac{b^2-1}{2^\gamma}$ and $\frac{a^2-1}{2^\gamma}$ are odd integers. Hence, there exist $i,j\in\{-1,+1\}$ and odd $u,v\in\mathbb{N}$ such that $a=2^{\gamma-1}u+i$ and $b=2^{\gamma-1}v+j$. As $c=ab-2^\gamma$, $c=2^\gamma w+k$ for some $w\in\mathbb{N}$ and $k:=ij$. From $$2^{\gamma}w\left(2^{\gamma-1}(iu+jv)+2\right)=(c-k)(ia+jb)=(ca-b)i+(bc-a)j=2^\beta i+2^\alpha j\,,$$ we conclude that $2^{\beta-\gamma-1}\mid w$ but $2^{\beta-\gamma}\nmid w$, or $c=2^{\beta-1}z+k$ for some odd $z\in\mathbb{N}$. Moreover, $$2^\beta=ca-b> ca-c=c(a-1)=\left(2^{\beta-1}z+k\right)(a-1) \geq \left(2^{\beta-1}-1\right)(a-1)$$ yields $a-1\leq 2$. Therefore, $a=3$, and this forces $z=1$ and $k=-1$. That is, $c=2^{\beta-1}-1$; i.e., $$2^\beta=ca-b=3\left(2^{\beta-1}-1\right)-b\,,$$ or $b=2^{\beta-1}-3$. Since $2^{\gamma-1}u+i=a$, we must have $\gamma=3$, $u=1$, and $i=-1$. Finally, $2^\gamma=ab-c$ implies that $8=3\left(2^{\beta-1}-3\right)-\left(2^{\beta-1}-1\right)=2^\beta-8$. That is, $\beta=4$, so $(a,b,c)=(3,5,7)$.
Solution 2:
A Sketch of a Possible Solution
Wlog order the solutions so that $a \ge b \ge c \ge 1$ with the system of equations
$\begin{align} ab - c &= 2^k \\ ac - b &= 2^l \\ bc - a &= 2^m \end{align}$
with $k \ge l \ge m \ge 0$.
We distinguish four cases:
- $a=b=c$
- $a=b>c$
- $a>b=c$
- $a>b>c$
Case 1: $a=b=c$
The main system of equations reduce to the single equation
$a(a-1) = 2^k \tag{1A}$.
$a,a-1$ cannot both be powers of 2 unless $a=2$. Therefore, the only solution for this case is
$\boxed{a=b=c=2}$
Case 2: $a=b>c$
Main equations reduce to
$\begin{align} a^2 - c &= 2^k \tag{2A}\\ a(c - 1) &= 2^l \tag{2B} \end{align}$
with $a > c, k > l$. Equation (2B) implies that $a$ is even since $a>c$, and (2A) now implies that $c$ is even since $k>l\ge0$. Then $a=2^l$ and so (2A) becomes: $2^{2l} = 2^k+2$ which can only be satisfied by $k=l=1$. However, this violates $k>l$ (it is just the first case again). No solutions for this case.
Case 3: $a>b=c$
Main equations reduce to
$\begin{align} b(a - 1) &= 2^k \tag{3A}\\ b^2 - a &= 2^m \tag{3B} \end{align}$
with $a > b, k > m$. Since $a>1$, (3A) implies $a$ is odd. On the other hand, (3B) implies $b>1$, so by (3A) $b$ must be even and $m=0$. Furthermore, we can put $a=2^p+1,b=2^q$ with $p\ge1,q\ge1$. So by (3B)
$b^2-a = 2^{2q}-(2^p+1) = 1$ so $2^{2q}-2^p = 2$. Whence $p=q=1$ must hold. So $a=3,b=2$ and the only solutions for this case are
$\boxed{a=3,b=c=2}$
Case 4: $a>b>c$
We restate the system of equations
$\begin{align} ab - c &= 2^k \\ ac - b &= 2^l \tag{4A}\\ bc - a &= 2^m \end{align}$
where we now have $a>b>c\ge1,k>l>m\ge0$. If exactly one of $a,b,c$ is even, then $ab-c,ac-b,bc-a$ are all odd, which cannot satisfy the system of equations with $k>l>m$ (see Theo Bendit's argument).
So now distinguish three other cases:
- $a,b,c$ all even
- Two of $a,b,c$ are even
- $a,b,c$ all odd
Case 4.1: $a>b>c$ with $a,b,c$ all even
We write $a=2p,b=2q,c=2r$ with $p>q>r\ge1$. Then system (4A) can be re-expressed as
$\begin{align} 4pq - 2r &= 2^k \\ 4pr - 2q &= 2^l \\ 4qr - 2p &= 2^m \end{align}$
Since the left-hand sides are all integers, divide through by 2 to get:
$\begin{align} 2pq - r &= 2^{k-1} \\ 2pr - q &= 2^{l-1} \\ 2qr - p &= 2^{m-1} \end{align}$
Since $p>q>r$, only $r$ can be odd. But if this is so, $m=1$ and this is excluded since $2pq-r\ge2\cdot3\cdot2-1=11$. So $p,q,r$ are all even. Then write $p=2s,q=2t,r=2u$ with $s>t>u$. Then system (4A) becomes
$\begin{align} 4pq - r &= 2^{k-2} \\ 4pr - q &= 2^{l-2} \\ 4qr - p &= 2^{m-2} \end{align}$
Using similar arguments we can show that $s,t,u$ must all be even. Since these arguments can be repeated ad infinitum, we conclude that there are no solutions for Case 4.1.
Case 4.2: $a>b>c$ with two of $a,b,c$ even
In this case, exactly one of $ab-c,ac-b,bc-a$ are odd, so to satisfy (4A) it must be the smallest of these: $bc-a$. Hence we must have $a$ odd, and $b,c$ even. Furthermore, $m=0$. So in (4A) we must have
$bc - a = 1 \tag{4.2A}$ with the restriction that $b>c\ge2$.
Now substitute the equivalent expression for $a$ into (4A) to get
$\begin{align} b^2c - b - c &= 2^k \\ bc^2 - b -c &= 2^l \tag{4.2B}\\ \end{align}$
Hence by subtraction
$\begin{align} b^2c - bc^2 &= 2^k - 2^l \\ bc(b-c) &= 2^l(2^{k-l}-1) \tag{4.2C}\\ \end{align}$
This is satisfied by $b=6,c=2$ or the solution $(a,b,c)=(11,6,2)$. There may be other solutions.