Question

Here is another one, (much easy than the earlier one)

How many real \( r \) are there such that the roots of \( x^2 + r x + 6r = 0 \) are both integers?

Answer 1

By the quadratic formula,
\[x={-r\pm \sqrt{r^2-24r} \over 2}\]
The roots are integers when \(r^2-24r=(r-12)^2-144\) is a complete square divisible by 4 and r is even.

Answer 2

perfect square I mean.

Answer 3

Yes, so how many solutions are there with that constraints?

Answer 4

There are infinitely many.

Answer 5

Assuming r is real as indicated in your question.

Answer 6

There is a finite number of integer solution.

Answer 7

That's not true. There are infinitely many r's for which \(r(r-24)\) is a perfect squares, but not all of them yield to an integer solution for x.

Answer 8

I mean what I said was not true @few :D

Answer 9

24 and 0 are obvious values for r that would give integer solutions.

Answer 10

I believe this is just a solution to the Pythagorean triples such that:

and the only such triples are:
12,12,0 => r-12=12
5,12,13 => r-12=13
9,12,15 => r-12=15
12,16,20 => r-12=20
12,35,37 => r-12=37
giving solutions of:
r=24,25,27,32,49
plus r=0 is a trivial solution

Answer 11

This is very nice, but someone has to be familiar with these numbers. Plus we need not to forget about negative values of r.

Answer 12

For example r=-3 is a solution.

Answer 13

So is r=-1.

Answer 14

true, negative r would give us:
|r-12|=12, 13, 15, 20, 37
giving us the following additional solutions:
r=-1, -3, -8, -25

Answer 15

Yeah! I should have seen it :D

Answer 16

@asnaseer:Magnum opus! Indeed Pythagorean triples are the only easy way out.

Answer 17

thx - you're teaching me Latin now :-D

Answer 18

Auf Wiedersehen ;)