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. ?

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Answer 1 · 2016-09-03T18:24:41

See below.

Supposing that

$(a^2+b^2)/(a b + 1) = k$ then

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

$a = (k bpm sqrt(k^2b^2-4(b^2-k)))/2$

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

$a=kb$ 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 aa and bb be positive integers such that ab + 1ab + 1 divides a^2 + b^2a^2 + b^2. Show that (a^2 + b^2) / (ab + 1)(a^2 + b^2) / (ab + 1) is the square of an integer. ?