equation on the set of integers:

**perash** · December 21st 2007, 11:59 PM

Solve the following equation on the set of integers:

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

**CaptainBlack** · December 22nd 2007, 02:26 AM

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

**Soroban** · December 22nd 2007, 04:53 AM

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.

**JaneBennet** · December 22nd 2007, 06:09 AM

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

**topsquark** · December 22nd 2007, 10:19 AM

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

**JaneBennet** · December 22nd 2007, 11:21 AM

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

**topsquark** · December 22nd 2007, 11:26 AM

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

**topsquark** · December 22nd 2007, 06:08 PM

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

Thread: equation on the set of integers:

LinkBack

Thread Tools

Display

equation on the set of integers:

Similar Math Help Forum Discussions

Find all positive integers for an equation

number of pair of integers (x,y) satisying an equation

equation with integers

positive integers equation

equation with integers

Search Tags