Finding all possible integer values of $a,b,c$, given $2$ equations. If $a^2=bc+1$ and $b^2=ac+1$, how do I find all possible values of $a,b,c$. Assuming $a,b,c$ all are integers?
 A: Subtract one equation from the other. Then
$$a^2 - b^2 = bc - ac$$
$$(a + b)(a - b) = -(a - b)c$$
$$(a + b)(a - b) + (a - b)c = 0$$
$$(a - b)(a + b + c) = 0$$
Considering the case where $a = b$: We end up with $a^2 = ac + 1$ so that $a(a - c) = 1$. Since $a, c$ are integers, we only have to consider two cases: $a, a-c = \pm1$.
Considering the case where $a + b + c = 0$: We get $c = -a - b$ so that
$$a^2 = b(-a - b) + 1$$
$$a^2 = -ab - b^2 + 1$$
$$a^2 + b^2 + ab - 1 = 0$$
$$a^2 + ba + b^2 - 1 = 0$$
Viewing this as a quadratic equation in $a$, we know that its discriminant must be nonnegative for real solutions of $a$ to exist. That is, $$b^2 - 4(b^2 -1)\geq0$$
$$-3b^2 + 4 \geq 0$$
$$3b^2 \leq 4$$
and using a similar argument,
$$3a^2 \leq 4$$
Here, we also only have to consider a few cases : namely only $b = 0, \pm 1, a = 0,\pm1$. Note that each possible pair of $a,b$ should be checked against the original equation $a^2 + ab + b^2 -1=0$.
A: The case when $|a| \leq 1$ or $|b| \leq 1$ can be solved by examination.
Otherwise we have that $a$ and $b$ have the same sign. If $(a,b,c)$ is a solution, so is $(-a,-b,-c)$, so we can assume w.l.o.g that  $0< a \leq b$.
Now the first equation implies: $a^2 \geq ac+1$ and the second implies: $ac+1 \geq a^2$. Thus $ac+1=a^2$, which again leads to $|a|=1$.
A: A SCHOOL QUESTION ABOUT THIS POST.
It follows $$ a ^ 2-b ^ 2 = (a-b) (a + b) = (a-b) (- c) $$ When $ a= b$ the integer solutions are only $ (a, b, c) = (\pm 1, \ ± 1,0) $.
When $ a\neq b$ we can simplify the factor $(a-b)$ so we get $$ a + b + c = 0 $$ Therefore we look at the integer points of the plan of equation $ X + Y + Z = 0 $ which have infinitely many of them. 
What is wrong with this reasoning?
