# Let a and b be positive integers such that ab + 1 divides a^2 + b^2. Show that (a^2 + b^2) / (ab + 1) is the square of an integer. ?

Sep 3, 2016

See below.

#### Explanation:

Supposing that

$\frac{{a}^{2} + {b}^{2}}{a b + 1} = k$ then

${a}^{2} - k b a + {b}^{2} - k = 0$ solving for $a$

$a = \frac{k b \pm \sqrt{{k}^{2} {b}^{2} - 4 \left({b}^{2} - k\right)}}{2}$

all the $b$'s obeying ${b}^{2} = k$ and the $a$'s obeying

$a = k b$ are solutions. So

$\left(\begin{matrix}b & a & k \\ 1 & 1 & 1 \\ 2 & 8 & 4 \\ 3 & 27 & 9 \\ 4 & 64 & 16 \\ 5 & 125 & 25 \\ 6 & 216 & 36 \\ \cdots & \cdots & \cdots\end{matrix}\right)$

The question follows. Are those all solutions?