1. ## find integer roots

thank you very much

Prove that if p,q,r and s are odd integers, then

x^10 + px^9 - qx^7 + rx^4 - s = 0

has no integer roots

2. Hello, ieatfood!

Prove that if $p,\,q,\,r,\,s$ are odd integers,
then: $x^{10} + px^9 - qx^7 + rx^4 - s \;= \;0$ has no integer roots.

[1] The sum of the ten roots is $-p$, an odd integer.
[2] The product of the ten roots is $s$, an odd integer.

From [2] . . . Since $s$ is odd, all ten roots must be odd integers.

But the sum of ten odd integers is even . . . which contradicts [1].

3. Originally Posted by Soroban
...

[1] The sum of the ten roots is $-p$, an odd integer.
[2] The product of the ten roots is $s$, an odd integer.

...

How do we know this again?

4. Originally Posted by Soroban
Hello, ieatfood!

[1] The sum of the ten roots is $-p$, an odd integer.
[2] The product of the ten roots is $s$, an odd integer.

From [2] . . . Since $s$ is odd, all ten roots must be odd integers.

But the sum of ten odd integers is even . . . which contradicts [1].
So then q and r don't matter? I was wondering about these.

-Dan

5. Originally Posted by topsquark
... I was wondering about these.

-Dan
Thank God i'm not the only one! I felt stupid asking that question

we stand together, topsquark

6. Originally Posted by Jhevon
Thank God i'm not the only one! I felt stupid asking that question

we stand together, topsquark
Yeah, but I know where Soroban got the theorem from...

I just don't know the name of it. (People keep telling me, but I always forget.)

-Dan

7. I'd be happy if one of you could provide us with some information about it

(How awesome are these Skype smilies!)

8. In the expansion of $\prod\limits_{k = 1}^N {\left( {x - r_k } \right)} = 0$ the coefficient of $x^{N - k} ,\;k = 1,2, \cdots N$ is $\left( { - 1} \right)^k S_k$.
Where $S_k$ is the sum of the products of the roots taken k at a time.

9. Originally Posted by Soroban
Hello, ieatfood!

[1] The sum of the ten roots is $-p$, an odd integer.
[2] The product of the ten roots is $s$, an odd integer.

From [2] . . . Since $s$ is odd, all ten roots must be odd integers.
And why must there be 10 real roots?

As far as I can see there are at most 6 real roots (but there must be an even number of them).

RonL

10. hey guys

im really desperate with this question..and i still dont understand it....PLEASE help!!!!! how do you know that there are 10 roots??im confused

thank you again

11. If you take the polynomial $f(x) = (x-x_0)(x-x_1)...(x-x_n)$ and multiply them out you get as coefficients (without signs) the so-called "symettric functions":
$x_0+x_1+...+x_n$
$x_0x_1+x_0x_2+...+x_{n-1}x_n$ (all possible products)
...
$x_0x_1...x_n$

The important thing about these is that they are invariant under and premutation of the roots.

This is the (unsigned) version of the theorem Soroban used.