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

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Answer 1 · 2016-09-03T18:24:41

See below.

Supposing that

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

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

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

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

$a = k b$ $a=kb$ are solutions. So

$((b,a,k),(1,1,1),(2,8,4),(3,27,9),(4,64,16),(5,125,25),(6,216,36),(cdots,cdots,cdots))$

The question follows. Are those all solutions?

Let aaa and bbb be positive integers such that ab + 1ab+1ab + 1 divides a^2 + b^2a2+b2a^2 + b^2. Show that (a^2 + b^2) / (ab + 1)a2+b2ab+1(a^2 + b^2) / (ab + 1) is the square of an integer. ?