Prove that $q$ is a perfect square

Question

Problem

If $a$, $b$, $q=\left(\dfrac{a^2+b^2}{1+ab}\right)$ are natural numbers, then $q$ is a perfect square. Are there infinitely many $\left(a,b,q\right)$ triplets?

I have tried to prove the problem assuming that $q$ is not a perfect square and hoped to get some contradiction but unfortunately I can't seem to get it. Regarding the second part of the problem, I think that probably it is not true. In case it is true I add the following problem to the original statement.

Can there be found expressions for $a$, $b$ and $q$ such that it generates an infinitely many (but not necessarily all) such triplets?

Any help will be appreciated.

See also http://www-groups.dcs.st-and.ac.uk/~john/Zagier/Problems.html. — lhf, Dec 02 '14 at 14:27
I wasn't aware of it. Thanks for the link. But What about the number of examples? — , Dec 02 '14 at 14:28
Since you are looking for $q\in\mathbb{Z}$, you need to determine when $1+ab|a^2+b^2$. — KSmarts, Dec 02 '14 at 14:36
natural numbers are positive integers: $1,2,3,4,...$. By definition, that excludes peterwhy's example. If domain was integers, then he has a perfect counter-example. — FundThmCalculus, Dec 02 '14 at 15:00

KSmarts · Answer 1 · 2014-12-02T15:30:27.950

0

Let $a=0$. Then $$ q=\frac{a^2+b^2}{1+ab}=\frac{0+b^2}{1+0}=b^2 $$ Also, since $b\in\mathbb{Z}$, you know that $q\in\mathbb{Z}$ and that $q$ is a perfect square. Therefore, for any $x\in\mathbb{Z}$, the triplet $(0,x,x^2)$ satisfies the requirements. So does $(x,0,x^2)$.

There are others, too, but you don't need to find them, since there are infinitely many already.

Restricting $a,b,q>0$, as per your edit, makes things a bit trickier. I'm still working on it, but I've only found the rather trivial $(1,1,1)$ so far.

edited Dec 02 '14 at 15:30

answered Dec 02 '14 at 14:55

KSmarts

3,364

I have edited the question. – Dec 02 '14 at 14:56
@user170039 I have changed my answer accordingly. It doesn't really affect much. – KSmarts Dec 02 '14 at 15:02
@user170039 Wait... are you considering $0$ to be a natural number or not? – KSmarts Dec 02 '14 at 15:02
Of course $0$ is not a Natural Number. – Dec 02 '14 at 15:03

Prove that $q$ is a perfect square

1 Answers1