Hi I have been working on this problem for a while and don't understand the solution given.
Here is the solution:
Suppose that $(a^2 + b^2)/(ab + 1) = k$ and we assume that there exists one or more solution to the given condition where k is not a perfect square. For a given value of k, let (A,B) be the solution to this equation that minimizes the value of A+B and WLOG, lets say that $A \ge B$. Let's look at the universe of solutions for (x,B).
$(a^2 + b^2)/(ab+1) = k$
$x^2 + B^2 = k(xB+1)$
$x^2 - kB(x) + (B^2 - k) = 0$
One root of the above equation is clearly A, and by vieta's formulas, the sum of the roots is $kb$ and the product of the roots is $b^2 - k$. Thus, the other root can be written as $c = (kb - A) = (b^2 - k)/A$. The left side of the equality means that the other root is an integer. Additionally, we know that this new root cannot be 0 because $b^2 = k$, and we assumed that k is not a perfect square. If $c < 0$, then $c^2 - kb(c) + (b^2 - k) \ge c^2 + k - b^2 - k > 0$.
The above inequality contradicts the fact that c has to be the root of the equation. Since we know that $A \ge B$, $c = (B^2 - k)/A < B^2/A < B < A$, thus implies that $c + B < A + B$.
Which contradicts the minimality of (A, B)
I don't get how they came to the left side of the equality means that the other root is an integer. I also don't get how they came to if $c < 0$, then $c^2 - kb(c) + (b^2 - k) \ge c^2 + k - b^2 - k > 0$. Why add the extra k's to $c^2 + k - b^2 - k$? And how does that inequality contradict c being a root?
Thanks!
