<s>[INST] A generalization of the (in)famous IMO 1988 problem 6: If $\frac{a^2 + b^2 - abc}{ab + 1}$ is a positive integer then it is a square. This question is motivated by the famous IMO $1988$ problem $6$. Is the following true? 

Let $a,b$ be positive integers and $c \ge 0$ be a non-negative integer. If $\dfrac{a^2 + b^2 - abc}{ab + 1}$ is a positive integer then it is a square. 

 [/INST] Yes, this fact is true. We have:
$$\frac{a^2+b^2-abc}{ab+1}=k \implies a^2-b(c+k)a+(b^2-k)=0$$
Following the standard procedure for Vieta jumping, assume that $(a,b)$ is the smallest solution with respect to the sum of solutions $a+b$, such that $k$ is not a perfect square. If one root of $$x^2-b(c+k)x+(b^2-k)$$ is $a$, then:
$$x=b(c+k)-a$$
$$x=\frac{b^2-k}{a}$$
WLOG let $a>b$. We can note that the first equation gives $x$ is an integer, while the second gives:
$$x<\frac{b^2}{a}<\frac{a^2}{a}=a \implies x<a$$
By the minimality of the solution $(a,b)$, we cannot have $(b,x)$ as a solution. Thus, we must have $x$ to be non-negative. We cannot have $x=0$ as this would give $b^2=k$ contradicting the assumption that $k$ is not a square. 
If $x<0$, we must have $b^2<k$. This gives:
$$(k-b^2)(ab+1)=a^2+b^2-abc-ab^3-b^2=a(a-bc-b^3) \implies \frac{k-b^2}{a}=\frac{a-bc-b^3}{ab+1}$$
$$-x=\frac{a-bc-b^3}{ab+1}<\frac{a}{ab+1}<1 \implies x>-1$$
This is a contradiction since there is no integer $0>x>-1$. Thus, there were no solutions to begin with, where $k$ is not a square.
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