Finding $(a, b, c)$ with $ab-c$, $bc-a$, and $ca-b$ being powers of $2$

Question

This is a 2015 IMO problem. It seems difficult to solve.

Find all triples of positive integers $(a, b, c)$ such that each of the numbers $ab-c$, $bc-a$, and $ca-b$ is a power of $2$.

Four such triples are $(a,b,c)=(2,2,2),(2,3,2),(3,5,7),(2,6,11)$.

Answer 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 $(a,b,c)=(3,5,7)$.

Answer 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:

1. $a=b=c$
2. $a=b>c$
3. $a>b=c$
4. $a>b>c$

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:

1. $a,b,c$ all even
2. Two of $a,b,c$ are even
3. $a,b,c$ 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.

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Answer 3

Due to symmetry, we can also consider $a\geq b\geq c$ with $ab-c=2^k, bc-a=2^l, ca-b=2^m$.

We will initially assume that $k,l,m>0$ with $k\geq m\geq l$.

Subtracting the third from the first, we get: $ab-c-ca+b=2^m(2^{k-m}-1)\Leftrightarrow (a+1)(b-c)=2^m(2^{k-m}-1) \ (1)$.

We will distinguish $2$ cases:

$\bullet$ Then, if $b=c$, it will be true $ab-b=2^k\Leftrightarrow b(a-1)=2^k$.

If $a-1$ is odd, it must be $1$, resulting in $a=2$ and $b=c=2^k$. With substitution in the second equation, we get $b^2=2^l+2$, which leads to $2^{2k}=2^l+2$, $l\leq k$. So, $2^{2l}-2^l-2\leq 0\Leftrightarrow (2^l-2)(2^l+1)\leq 0$. Therefore, $2^l=1$ (rejected) or $2^l=2$. In this case, we get the solution $\boxed{(a,b,c)=(2,2,2)}$.

Then, $b\neq c$, since $a+1$ is superfluous, it holds $2^m\mid b-c$. Therefore, $2^m\leq b-c\Leftrightarrow ca-b\leq b-c\Leftrightarrow c(a+1)\leq 2b$ (due to the chosen layout), which is possible only if $c=1$, leading to no solutions.

$\bullet$ Suppose $a$ is superfluous; the other $a+1,a-1$ can be divisible at the same time by $4$.

Suppose $4\nmid a+1$. Then, according to $(1)$, $2^{m-1}\mid b-c$. If $b=c$, then $ab-b=2^k\Leftrightarrow b(a-1)=2^k$.

For $b=1$, there are no solutions. For $b=2^n$, $b^{2}-2=2^{2n}-2$. But $b^2-2=2^p$, so $2^{2n}-2^p=2$, yielding $2$ forces that make a difference. So, $b=2$ leads to the solution $\boxed{(a,b,c)=(3,2,2)}$, but in the first instance, it is rejected since we assumed that $4\nmid a+1$.

Therefore, $b\neq c$, and $2^{m-1}\mid b-c$. If $b\neq c$, then $2^{m-1}\mid b-c$, $b-c\geq \frac{ca-b}{2}\Leftrightarrow 3b\geq c(a+2)$ (due to the provision we have chosen), so $c\leq 2$.

If $c=1$, there are no solutions; if $c=2$, the second equation gives $2b-a=2^l$. i.e., $a$ would be even, which is out of place.

Let us now assume that $4\nmid a-1$. It is superfluous $bc=2^l+a$, and therefore superfluous $b,c$. Adding the first to the third ratio, we get $(a-1)(b+c)=2^m(2^{k-m}+1) \ (2)$.

It will necessarily be the case $2^{m-1}\mid b+c$. $b+c\geq \frac{ca-b}{2}\Leftrightarrow c(a-2)\leq 3b$ due to the provision we have chosen, it follows that $c\leq 5$.

$\bullet$ If $c\geq 5$, then $5(a-2)\leq c(a-2)\leq 3b\Leftrightarrow 5a-3b\leq 10\overset{a\geq b}\Leftrightarrow 2b\leq 5a-3b\leq 10\Leftrightarrow b\leq 5$. So, it would be true $b<c$, which is out of place. Therefore, $c\leq 5$.

$\bullet$ If $c=5$, then identical to before, we end up with the solution $a=b=c=5$, which is not acceptable.

$\bullet$ If $c=1$, then obviously, there are no solutions.

$\bullet$ If $c=3$, then $ab=2^k+3$ and $2b-a=1$, $2a-b=2^m$. If both are redundant, then from the third relationship, we get $2^m$ redundant, so $2a-b=1$. But $a=b=1$ is also true, but we do not get solutions from it.

If any of them is even, then $ab$ is even, therefore $2^k+3$ is even, therefore $k=0$. But then it necessarily applies (because of the provision) and $m=0$; therefore, we get again $a=b=1$, which is out of place. .

$\bullet$ If $c=4$, then $ab=2(2^{k-1}+2)$ and $4b-a=1$, $4a-b=2^m$. Because $a=4b-1$, the second relationship is written $16b-4-b=2^m\Leftrightarrow 15b=2^m+4$. Obviously, $m\neq 0$, we take it $\mod 3$ that the one is superfluous.

Let $m=2n+1$. We then $\mod 5$ that $2^{m}+4\equiv 2\cdot 4^{n}-1\equiv 2\cdot (-1)^{n}-1\pmod 5$. This cannot be equal to $0\pmod 5$, which we saw above is inappropriate $2^{m}+4=15b\equiv 0\pmod 5$.

$\bullet$ If $c=2$, then $ab=2^k+2$ and $2b-a=1 \ (3)$ and $2a-b=2^m \ (4)$.

From the last two, we $b=\frac{2^m+2}{3}$ and $a=\frac{2(2^m+2)}{3}-1$. Substituting in the former and doing deeds, we arrive at $2^{2m+1}+5\cdot 2^m=9\cdot 2^{k}+16$.

Because $k\geq m$, we can write $2^{m+1}+5=9\cdot 2^{k-m}+2^{4-m}$. Because $2^{m+1},5,2^{k-m}$ are integers, must also be $2^{4-m}$ an integer.

So, $m\leq 4$. By checking the cases (all very simple), we also get the solutions $\boxed{(a,b,c)=(3,2,2),(11,6,2)}$.

Therefore, $\boxed{(a,b,c)=(2,2,2),(3,2,2),(11,6,2),(7,5,3)}$, together with all their permutations, they are all solutions to the problem.

Answer 4

Assume a,b,c satisfy the hypothesis.

For simplicity, also assume a <= b <= c.

To restore symmetry, as solution triples are found, we will  permute them in all possible ways.

If a = 1 then

ab - c = b - c <= 0,

which is impossible since ab - c is a power of 2.

Hence a >= 2.

Let x = ab - c, y = ac - b, z = bc - a.

Then x,y,z are powers of 2 with x <= y <= z.

As a consequence, x|y, x|z, y|z.

Then y = ac - b >= a^2 - a = a(a - 1) >= 2 [since a >= 2], hence since y is a power of 2, y is even.

Thus, y,z are both even powers of 2.

Consider the following 4 cases:

Case (1): a = b = c Case (2): a = b, b < c Case (3): a < b, b = c Case (4): a < b < c

Case (1): a = b = c

Then 

x is a power of 2

=> a^2 - a is a power of 2

=> a(a - 1) is a power of 2

=> a and a - 1 are both powers of 2

=> a = 2

=> a = b = c = 2

=> x = y = z = 2

so we have the solution (a,b,c) = (2,2,2).

Case (2): a = b, b < c

Then 

x = a^2 - c

=> a > 1

Also,

y = ac - a = a(c - 1)

=> a and c - 1 are both powers of 2.

Since a >= 2, a must be an even power of 2.

Also

a < c

=> c >= 3 

=> c - 1 is an even power of 2

=> c is odd

=> x is odd [since x = a^2 - c]

=> x = 1 [since x is a power of 2]

=> a^2 - c = 1

=> c = a^2 - 1

=> c - 1 = a^2 - 2

=> c - 1 = 2 (mod 4)

=> c - 1 = 2 [since c - 1 is a power of 2]

=> c = 3

Thus, we have the candidate solution (a,b,c) = (2,2,3).

For (a,b,c) = (2,2,3) we get

x = ab - c = 1

y = ac - b = 4

z = bc - a = 4

so allowing for permutations, we get the 3 solutions

(a,b,c) = any permutation of (2,2,3)

Case (3): a < b, b = c

Then

x = ab - c = ab - b = b(a - 1)

y = ac - b = ab - b = b(a - 1)

z = bc - a = b^2 - a

Thus, b and a - 1 must both be powers of 2.

Since b > a - 1, b is an even power of 2.

Since b and z are both even, 

z = b^2 - a 

=> a is even

=> a - 1 is odd

=> a - 1 = 1 [since it's a power of 2]

=> a = 2

=> z = 2 (mod 4)

=> z = 2 [since z is a power of 2]

=> b^2 - a = 2

=> b^2 = 4

=> b = 2, contrary to a < b

Case (4): a < b < c

y|z => y|(z + y) and y|(z - y).

Simplifying,

z + y = (bc - a) + (ac - b) = (b + a)(c - 1)

z - y = (bc - a) - (ac - b) = (b - a)(c + 1)

Hence 

y|(b + a)(c - 1)

y|(b - a)(c + 1)

But c - 1 and c + 1 cannot both be multiples of 4, hence since y is an even power of 2, it follows that either

(y/2)|(b + a) or (y/2)|(b - a).

Then

(y/2)|(b + a) or (y/2)|(b - a)

=> y/2 <= b + a

=> (ac - b)/2 <= b + a

=> ac <= 3b + 2a

=> a(b + 1) <= 3b + 2a

=> ab + a <= 3b + 2a

=> b(a - 3) <= a

=> a - 3 <= 0

=> a <= 3

Hence a = 2 or a = 3.

Consider the following 2 subcases:

First suppose a = 2. Then

x = 2b - c

y = 2c - b

z = bc - 2

But then

a < b < c 

=> x < y < z 

=> 4|z 

=> 4|(bc - 2) 

=> one of b,c is even, the other odd.

Since y is even, so is b, hence b is even, c is odd.

Since c is odd, x is odd, hence x = 1. Then

x = 1

=> 2b - c = 1

=> c = 2b - 1

=> y = 3b - 2 and z = 2b^2 - b - 1

Then 

a < b => b >= 3 => b >= 4 [since b is even],

so y = 3b - 2 >= 10

Identically,

9z = y(6b - 1) - 16

hence y|z => y|16, but then, since y >= 10, y = 16.

Then 3b - 2 = 16 => b = 6, and c = 2b - 1 => c = 11.

Thus, we have the candidate solution (a,b,c) = (2,6,11).

For (a,b,c) = (2,6,11) we get

x = ab - c = 1

y = ac - b = 16

z = bc - a = 64

so allowing for permutations, we get the 6 solutions:

(a,b,c) = any permutation of (2,6,11)

This completes the analysis for the subcase a = 2.

Next suppose a = 3. Then

x = 3b - c

y = 3c - b

z = bc - 3

But then

a < b < c 

=> x < y < z 

=> 4|z 

=> 4|(bc - 3) 

=> b,c are both odd and one of b,c = 1 (mod 4), the other = 3 (mod 4).

In either case, 

x = 3b - c = 0 (mod 4)

Then

x|y

=> x|(3x + y)

=> x|(8b) 

=> x|8 [since b is odd]

=> x = 4 or 8 [since 4|x]

First suppose x = 4. Then

x = 4

=> 3b - c = 4

=> c = 3b - 4

But then

y = 3c - b

=> y = 3(3b - 4) - b

=> y = 8b - 12

=> y = 4(2b - 3)

=> 2b - 3 is a power of 2

=> 2b - 3 = 1

=> b = 2, contradiction since b is odd

Hence we must have x = 8. Then

x = 8

=> 3b - c = 8

=> c = 3b - 8

Next, look at y ...

y = 3c - b

=> y = 3(3b - 8) - b

=> y = 8b - 24

=> y = 8(b - 3)

=> b - 3 is a power of 2

=> b - 3 is an even power of 2 [since b is odd]

=> b - 3 >= 2

=> b >= 5

Also, look at z ...

z = bc - 3

=> z = b(3b - 8) - 3

=> z = 3b^2 - 8b - 3 = (3b + 1)(b - 3)

hence

y|z

=> (8(b - 3))|((3b + 1)(b - 3))

=> 8|(3b + 1)

=> 3b = -1 (mod 8)

=> b = 5 (mod 8)

But then, since z is a power of 2,

z = (3b + 1)(b - 3) 

=> 3b + 1 and b - 3 are both powers of 2 [since z is]

=> (3b + 1)/(b - 3) is a power of 2

=> 3 + 8/(b - 3) is a power of 2

=> b - 3 <= 8

=> b <= 11

=> b = 5 [since b = 5 (mod 8)]

=> c = 3b - 8 = 7

Thus, we have the candidate solution (a,b,c) = (3,5,7).

For (a,b,c) = (3,5,7) we get

x = ab - c = 8

y = ac - b = 16

z = bc - a = 32

so allowing for permutations, we get the 6 solutions:

(a,b,c) = any permutation of (3,5,17)

This completes the analysis for the subcase a = 3.

At this point, we have all the solutions:

(a,b,c) = (2,2,2)

(a,b,c) = any permutation of (2,2,3)

(a,b,c) = any permutation of (2,6,11)

(a,b,c) = any permutation of (3,5,7)

This completes the analysis.

Answer 5

Without loss of generality: replace $b$ with $a+k1$ and $c$ with $a+ k2$

$(a, a+k1, a +k2) $

$ab- c = a2 + k1a -a -k2 $

$bc- a = a2 + k1k2 + (k1+k2 -1)a$

$ca- b = a2+ (k2-1)a - k1$

All these must by $2N$ for some $n$ in $I > 0$