Mad for solutions

We have that

$a^{2} - b^{2} = (bc + 1) - (ca + 1) = c*(b - a) \Longrightarrow (a - b)(a + b) = -c*(a - b)$

$\Longrightarrow (a - b)(a + b + c) = 0.$

So either (i) $a - b = 0$ or (ii) $a + b + c = 0$ .

(i) If $a - b = 0 \Longrightarrow a = b$ then $a^{2} = ca + 1 \Longrightarrow c = \dfrac{a^{2} - 1}{a} = a - \dfrac{1}{a}$ for $a \ne 0.$

(Note that $a = b = 0$ is not a possible solution as the original equations would then both yield the absurdity $0 = 1$ .)

In this case $c$ will be an integer only if $a = 1$ or $a = -1$ . The respective solution triplets $(a,b,c)$ are then $(1,1,0)$ and $(-1,-1,0)$ .

(ii) If $a + b + c = 0 \Longrightarrow c = -(a + b)$ then $a^{2} = bc + 1 = -b(a + b) + 1$

$\Longrightarrow a^{2} + ba + b^{2} - 1 = 0 \Longrightarrow a = \dfrac{-b \pm \sqrt{b^{2} - 4(b^{2} - 1)}}{2} = \dfrac{-b \pm \sqrt{4 - 3b^{2}}}{2}.$

Now $a$ will be an integer only if $4 - 3b^{2} \gt 0$ , which is only the case when $b = 0, \pm 1$ .

When $b = 0$ we have $a = \dfrac{0 \pm\sqrt{4}}{2} = \pm 1$ , yielding solution triplets $(1,0,-1)$ and $(-1,0,1)$ .

When $b = 1$ we have $a = \dfrac{-1 \pm \sqrt{4 - 3}}{2} = 0,-1$ , yielding solution triplets $(0,1,-1)$ and $(-1,1,0)$ .

When $b = -1$ we have $a = \dfrac{1 \pm \sqrt{4 - 3}}{2} = 1,0$ , yielding solution triplets $(1,-1,0)$ and $(0,-1,1).$

Having considered all possible cases, we see that the total number of desired solution triplets is $\boxed{8}.$

Solve for $c$ to get $c = \frac{a^2-1}{b} = \frac{b^2-1}{a}$ . So $a^3-a = b^3-b$ .

First: if $a = b$ , then we get $a(a-c) = 1$ , so $a = b = \pm 1$ . We'll come back to this case in a second.

Now look at a graph of the function $y = x^3-x$ . It's increasing and negative until $(-1,0)$ and increasing and positive after $(1,0)$ . The only horizontal line that passes through two points with integer $x$ -coordinate is $x = 0$ , which passes through the points where $x = -1,0,1$ . So if $a \ne b$ , then $a$ and $b$ have to be in $\{ -1,0,1 \}$ .

So in either case $a$ and $b$ are in $\{ -1,0,1 \}$ . So there are nine potential solutions, and it's easy to compute that the only possibility that doesn't lead to a solution is $a = b = 0$ . So there are $\fbox{8}$ solutions in all.

This question was in rmo 2015

a^2=bc+1

.•. (a+1)(a-1)=bc

Similarly, (b+1)(b-1)=ac

Dividing both the equations we get

[(a+1)(a-1)]/[(b+1)(b-1)]=b/a

.•. (a-1)(a)(a+1)=(b-1)(b)(b+1)

Here we get two cases

(i) a=b

Then,

a^2=bc+1

But since a=b

.•. a^2=ac+1

.•. c=(a^2-1)/a

Which will give you an integer only if a=1 or -1

Thus, we get two solutions (1,0,1) and (-1,0,-1)

(ii) a≠b

Then,

For (a-1)(a)(a+1)=(b-1)(b)(b+1)

To be true both quantities should have one multiplication by zero.

.•. taking

1) (a-1)=b=0

2) (a-1)=(b+1)=0

3) a=(b-1)=0

4) a=(b+1)=0

5) (a+1)=(b-1)=0

6) (a+1)=b=0

By these six cases we get the value of a and b. We can find c by substituting in original equation.

We get the six permutations of (-1,0,1).

Which give us a total of 8 solutions.