need some help pleasee
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
thank you in advance
Hello, ieatfood!
Prove that if $\displaystyle p,\,q,\,r,\,s$ are odd integers,
then: $\displaystyle x^{10} + px^9 - qx^7 + rx^4 - s \;= \;0$ has no integer roots.
[1] The sum of the ten roots is $\displaystyle -p$, an odd integer.
[2] The product of the ten roots is $\displaystyle s$, an odd integer.
From [2] . . . Since $\displaystyle s$ is odd, all ten roots must be odd integers.
But the sum of ten odd integers is even . . . which contradicts [1].
In the expansion of $\displaystyle \prod\limits_{k = 1}^N {\left( {x - r_k } \right)} = 0$ the coefficient of $\displaystyle x^{N - k} ,\;k = 1,2, \cdots N $ is $\displaystyle \left( { - 1} \right)^k S_k $.
Where $\displaystyle S_k$ is the sum of the products of the roots taken k at a time.
If you take the polynomial $\displaystyle f(x) = (x-x_0)(x-x_1)...(x-x_n)$ and multiply them out you get as coefficients (without signs) the so-called "symettric functions":
$\displaystyle x_0+x_1+...+x_n$
$\displaystyle x_0x_1+x_0x_2+...+x_{n-1}x_n$ (all possible products)
...
$\displaystyle x_0x_1...x_n$
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.