Polynomial with integer solutions.

**davismj** · March 1st 2010, 03:42 PM

Am I doing it correctly? What would the next step be?

**icemanfan** · March 1st 2010, 04:11 PM

If you let $y = x + k$ , then k must be an integer and x must be an integer solution to $x^2 + kx - 12k = 0$ . Hence, the discriminant of this polynomial, which is $k^2 + 48k = k(k + 48)$ , must be a square. Finally, let $k = n - 24$ . Then $(n - 24)(n + 24) = n^2 - 24^2$ must be a square. Now use your knowledge of pythagorean triples and multiples of them to find solutions:

$(3, 4, 5) \to (24, 32, 40); (18, 24, 30)$
$(5, 12, 13) \to (10, 24, 26)$
$(7, 24, 25)$
$(8, 15, 17) \to (24, 45, 51)$
$(9, 40, 41)$
$(11, 60, 61)$
$(12, 35, 37) \to (24, 70, 74)$
$(24, 143, 145)$

Now we need hypotenuses $n > 24$ . From these triples we get the solutions $25, 26, 30, 40, 51, 74, 145$ .

This yields the following solutions for k:
$1, 2, 6, 16, 27, 50, 121$

That gives you seven quadratic equations to solve for x, some of which may not yield integer solutions.

**davismj** · March 1st 2010, 04:30 PM

Hence, the discriminant of this polynomial must be a square.

Why?

Edit: Quadratic formula? Since we want integer solutions, the discriminant must be a perfect square?

Now use your knowledge of pythagorean triples and multiples of them to find solutions:

Sorry, but solutions to what? (n-24)(n+24)? or $n^2 = 24^2$ ? In either case, where do Pythagorean triples come into play?

Edit: Pythagorean Theorem. $n=24, (n)^2 - 24^2 = k^2 \implies k^2 + 24^2 = n^2.$ Is that about right?

Now we need n > 24.

I'm assuming the answer to this follows from question (2) above.

Edit: $k = n - 24, k > 0$ , so $n > 24$ . Although, shouldn't this go without saying since $n = 24, n^2 = 24^2 + k^2 \implies k = 0$ , which doesn't make sense?

Thanks so much for your help.

**icemanfan** · March 1st 2010, 04:51 PM

Sorry, that was a hopelessly convoluted method. You are looking for any value of n such that $n^2 - 24^2 = a^2$ for some integer a.

I am editing my post to take out values of n that don't work. Where the Pythagorean triples come in: n is the hypotenuse of a right triangle, with one side equal to 24, and the other side equal to a.

It is much easier to observe that x must be an integer from 1 to 11, and try each of those values and determine whether y ends up being an integer.

**davismj** · March 1st 2010, 05:09 PM

One quick question: Is there another way to solve this without the knowledge of pythagorean triples that I don't have?

Thanks

**Bacterius** · March 2nd 2010, 03:22 AM

Hello !

You want to solve $\frac{1}{x} + \frac{1}{y} = \frac{1}{12}$ for $x$ and $y$ . This is how I would have done :

$\frac{1}{x} = \frac{1}{12} - \frac{1}{y}$

$\frac{1}{x} = \frac{y}{12y} - \frac{12}{12y}$

$\frac{1}{x} = \frac{y - 12}{12y}$

$x = \frac{12y}{y - 12}$

All you have left is to determine for which values of $y$ do we have an integer $x$ , that is, when $(y - 12) | 12y$

Does it make sense ?

**davismj** · March 2nd 2010, 06:10 AM

Originally Posted by Bacterius

Hello !

You want to solve $\frac{1}{x} + \frac{1}{y} = \frac{1}{12}$ for $x$ and $y$ . This is how I would have done :

$\frac{1}{x} = \frac{1}{12} - \frac{1}{y}$

$\frac{1}{x} = \frac{y}{12y} - \frac{12}{12y}$

$\frac{1}{x} = \frac{y - 12}{12y}$

$x = \frac{12y}{y - 12}$

All you have left is to determine for which values of $y$ do we have an integer $x$ , that is, when $(y - 12) | 12y$

Does it make sense ?

Yes, but you copied the problem incorrectly.

**Bacterius** · March 2nd 2010, 09:34 AM

Oh, yeah. Well it doesn't actually change much, just replace the signs

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