Why does the one paragraph solution to IMO Problem 6 1988 work? Emanouil Atanassov, famously said to have completed the "hardest" IMO problem in a single paragraph and went on to receive the special prize, gave the proof quoted below,

Question: Let a and b be positive integers such that $ab+1$ divides $a^2+b^2$ Show that $\frac{a^2+b^2}{ab+1}$ is the square of an integer


Proof: $k=\frac{a^2+b^2}{ab+1} \implies a^2-kab+b^2=k, k\in \mathbb{Z}$ Assume $k$ is not a perfect square. Note that for any integral solution $(a,b)$ we have $a>0, b>0$ since k is not a perfect square. Let $(a,b)$ be an integral solution with $a>0, b>0$ and $a+b$ minimum. We shall produce from it another integral solution $(a',b)$ with $a'>0 , \ b>0$ and $a'+b<a+b$. Contradiction (We omit the argument for arriving at $(a',b)$)

$a'=0$ is sufficient for $k$ being a square, but it is not true in general. This proof seems to imply $a'=0$ for all solutions $(a,b)$. The only assumption contradicted is the minimality of $a+b$, not the assumption $k$ is not a perfect square. How does the assertion trivially follow from this proof?
EDIT:
Here is the proof modified, but without the assumption $k$ is not a perfect square.

$k=\frac{a^2+b^2}{ab+1} \implies a^2-kab+b^2=k, k\in \mathbb{Z}$ Let $(a,b)$ be an integral solution with $a>0, b>0$ and $a+b$ minimum. We shall produce from it another integral solution $(a',b)$ with $a'>0 , \ b>0$ and $a'+b<a+b$. Contradiction (We omit the argument for arriving at $(a',b)$)

I have also removed the second sentence, because $a,b>0$ is given in the question. What does this proof imply that the first does not?
 A: *

*If there are solutions $(a,b)$ for which $k$ is not a perfect square, then $a,b>0$.

*Also, if there are solutions $(a,b)$ for which $k$ is not a perfect square, then there will be, among those solutions, one for which $a+b$ is minimal.

*Then the author finds another solution $(a',b)$ with $a'<a$, which implies that $a'+b<a+b$.

*But that's impossible, since we were assuming that $(a,b)$ was the solution for which $a+b$ takes the smallest value.

A: Full solution verbatim from en.wiki/Vieta jumping:
Standard Vieta jumping
The concept of standard Vieta jumping is a proof by contradiction, and consists of the following three steps:${}^{[1]}$

*

*Assume toward a contradiction that some solution exists that violates the given requirements.

*Take the minimal such solution according to some definition of minimality.

*Show that this implies the existence of a smaller solution, hence a contradiction.

Example
Problem #6 at IMO 1988: Let $a$ and $b$ be positive integers such that $ab + 1$ divides $a^2 + b^2$.  Prove that $\frac{a^2 + b^2}{ab + 1}$ is a perfect square.${}^{[2]}$${}^{[3]}$

*

*Fix some value $k$ that is a non-square positive integer. Assume there exist positive integers $(a, b)$ for which $k = \frac{a^2 + b^2}{ab + 1}$.

*Let $(A, B)$ be positive integers for which $k = \frac{A^2 + B^2}{AB + 1}$ and such that $A + B$ is minimized, and without loss of generality assume $A \ge B$.

*Fixing $B$, replace $A$ with the variable $x$ to yield $x^2 – (kB)x + (B^2 – k) = 0$.  We know that one root of this equation is $x_1 = A$. By standard properties of quadratic equations, we know that the other root satisfies $x_2 = kB – A$ and $x_2 = \frac{B^2 – k}{A}$.

*The first expression for $x_2$ shows that $x_2$ is an integer, while the second expression implies that $x_2 \ne 0$ since $k$ is not a perfect square. From $\frac{x_2^2 + B^2}{x_2B + 1} = k > 0$ it further follows that $x_2$ is a positive integer. Finally, $ A \ge B$ implies that $x_2 = \frac{B^2 − k}{A} < A$ and thus $x_2 + B < A + B$, which contradicts the minimality of $A + B$.

A: I think I have it figured out and will allude to the Wikipedia proof given in Alexey's answer as the arguments are the same and I believe me source has been unreliable in "omitting" steps.
The minimality of $A+B$ is contradicted. (2) and (3) are irrelevant to $k$. (4) says $x$ cannot be $0$ if $k$ is not a perfect square. So $x\neq 0$. But if $x\neq 0$, purely through algebra, independent of $k$ being square or not, we contradict minimality. So, the crux, $(A,B)$ minimises $A+B$. only if $x_2=0$. Since there is no minimum of $(A,B)$ pairs when $k$ is not a square, we can conclude there are no such pairs.
Whether or not Atanassov found this so trivial that he kept this in his head remains a mystery.
