Recall the famous IMO 1988 question 6:
Suppose that $\displaystyle\frac{a^2+b^2}{ab+1}=k\in\mathbb{N}$ for some $a,b\in\mathbb{N}$. Show that $k$ is a perfect square.
Solutions can be found:
$1)$ http://projectpen.files.wordpress.com/2008/10/pen-vol-i-no-1.pdf (Section 2.6)
$2)$ http://www.georgmohr.dk/tr/tr09taltvieta.pdf
When I first saw solutions to this problem I had no reluctance in accepting the validity of solutions, but just recently I read over them again and there seems to be several things bugging me and it all relates to a particular step in all solutions above.
Solutions $1)$ and $2)$ are essentially the same argument in the sense that they both assume firstly that $k$ is not a perfect square and then deduce a contradiction.
Now let us get to the issue at hand. It follows from Vieta's formulas that the other root $c$ (the other root is called $a_1$, $x_2$ in $1)$, $2)$ respectively - I'm just going to call it $c$ for simplicity as a general reference to the other root of $x^2-kbx+b^2-k$ where $a$ is the first root) must be an integer. Now note that in proving that $c\ge 0$ for $1)$ and $2)$ we didn't need to use the assumption that $k$ is not a perfect square. But to show that $c\neq 0$ then we must finally use the assumption that $k$ is not a perfect square. Here is where there seems (at least to me at the time being) to an issue. Recall that in the problem we want $a,b\in\mathbb{N}$, so this is equivalent to defining a set where we fix $k$. $$S(k):= \left\{(a,b)\in\mathbb{N}\times\mathbb{N} : \frac{a^2+b^2}{ab+1}=k\in\mathbb{N}\right\}$$
Now in $1)$ and $2)$ they assume that $k$ is not a perfect square and hence deduce that $c\neq 0$ because if $c=0$ then as a result from $x^2-kbx+b^2-k=0$ then $k=b^2$. Here is where there seems to be an invalid contradiction. If $c=0$ then it doesn't contradict that $k=b^2$ because $c=0\not\in\mathbb{N}$ and hence $(c,b)\not\in S(k)$ so as a result $c^2-kbc+b^2-k=0$ no longer satisfies the condition of the question and hence doesn't contradict anything even though $k=b^2$. This is where there may be something I misunderstood and it would be great if someone can justify with clarity as to why my reasoning is flawed?
A little note: We see that it was deduced in $1)$ and $2)$ that $c$c\ge 0$ are all deduced without assuming $k$ not being a perfect square nor $b$ being minimal for fixed $k$.