Solve an equation with parameters over the naturals

Question

How can I find all natural numbers $a$ and $b$ so that the roots of the equation $x^2-abx+a+b$ are integer? ($x$ is a variable)

The first idea that came into my mind was to find the solutions of the equation. They are: $x = \frac{ab - \sqrt{a^2 b^2 - 4a - 4b}}{2}; x = \frac{ab + \sqrt{a^2 b^2 - 4a - 4b}}{2}$

What is possible to be done next?

Your title says "naturals" but the question says "integers". My answer was based on the title. — saulspatz, Feb 04 '18 at 20:19
It is naturals. I chose the tag integers because there is not tag "naturals" and it is the closest. — G. Amber, Feb 04 '18 at 20:21

saulspatz · Answer 1 · 2018-02-04T20:27:55.057

Suppose the roots are $c$ and $d$. Then $$(x-c)(x-d)=x^2-abx+a+b$$ gives $c+d=ab, a+b=cd$. Now we normally expect that $a+b < ab,$ s0 we have $$a+b<ab=c+d<cd,$$ contradicting $a+b=cd.$

I leave it to you to find the cases where $a+b < ab$ is violated and which ones give answers to the problem.

Your way of starting out was a fine idea, but it just doesn't happen to pan out, so far as I can see.

EDIT I wrote the above before the clarification about integers and naturals. I had assumed that $c$ and $d$ were naturals also. However, that's easy to deduce from $c+d=ab, a+b=cd$.

Solve an equation with parameters over the naturals

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