# Thread:

1. ## equation on the set of integers:

Solve the following equation on the set of integers:

$\sin(\frac{\pi}{3}(x-\sqrt{x^2-3x-12}))=0$

2. Originally Posted by perash
Solve the following equation on the set of integers:

$\sin(\frac{\pi}{3}(x-\sqrt{x^2-3x-12}))=0$
Which is asking you to find all the solutions of:

$x-\sqrt{x^2-3x-12}\equiv 0 \mod 3$

RonL

3. Hello, perash!

I haven't solve it yet . . . It's quite involved . . .

Solve the following equation on the set of integers:

$\sin\left[\frac{\pi}{3}\left(x-\sqrt{x^2-3x-12}\right)\right]\:=\:0$
The Captain is absolutely correct . . .

We have: . $\sin\left[\pi\left(\frac{x - \sqrt{x^2-3x-12}}{3}\right)\right] \:=\:0$

We know that: . $\sin(\pi k) \:=\:0$ . . . for any integer $k.$

Hence: . $\frac{x-\sqrt{x^2-3x-12}}{3}$ must be an integer.

. . We can let: . $\frac{x-\sqrt{x^2-3x-12}}{3} \:=\:k$

There are more restrictions:

. . $x^2-3x-12$ must be nonnegative: . $x^2-3x-12 \:\geq\:0\quad\Rightarrow\quad x \leq -3,\;x \geq 6$

. . And, of course, $x^2-3x-12$ must be a square.

4. So far I’ve found three possible values for x: −13, 7, 16.

5. Originally Posted by CaptainBlack
$x-\sqrt{x^2-3x-12}\equiv 0 \mod 3$
Okay, my stab at this is confusing me.

$x - \sqrt{x^2 - 3x - 12} \equiv 0~\text{ mod 3}$

$x - \sqrt{x^2} \equiv 0~\text{ mod 3}$

$x - |x| \equiv 0~\text{ mod 3}$

Isn't this an identity in mod 3? But the original equation is not true for all x within the domain.

-Dan

6. You can’t remove multiples of 3 within the square root. $\sqrt{4-3\times1}$ ≢ $\sqrt{4}\pmod{3}$.

7. Originally Posted by JaneBennet
You can’t remove multiples of 3 within the square root. $\sqrt{4+3\times7}$ ≢ $\sqrt{4}\pmod{3}$.
Hmmm... I had never noticed that before. Thank you.

(Upon reflection I should have known that. I studied that section of Algebra at one point.)

By the way, if you like, you can do the "not equivalent" in LaTeX: $\not \equiv$ by typing \not \equiv

-Dan

8. Originally Posted by JaneBennet
So far I’ve found three possible values for x: −13, 7, 16.
16 doesn't work.

I have "reduced" the problem to solving the Diophantine equation:
$n^2 - 4m^2 = 57$
(I can explain how I got this, but I'd like the original poster to do some work on their own.)

So
$(n + 2m)(n - 2m) = 57$

There are going to be some obvious limits on m and n here because n + 2m has to be one of 1, 3, 19, 57.

So we are looking the solution to one of two systems of equations:
$n + 2m = 57$
and
$n - 2m = 1$

or
$n + 2m = 19$
and
$n - 2m = 3$

The first has a solution of m = 14 and n = 29, giving an x value of -13. The second has a solution of m = 4 and n = 11, giving an x value of 7.

So the only two solutions are x = -13 and x = 7.

-Dan
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