# Divisibility

• Sep 22nd 2006, 10:57 AM
JaysFan31
Divisibility

If r is in Z and r is a non-zero solution of x^2+ax+b=0 (where a,b are in Z) prove that r divides b.

I solved it the following way:
Letting c and r be the roots of the polynomial.
(x-c)(x-r)=0
x^2+x(-r-c)+cr=0
cr=b, which is the definition of r dividing b

The question then goes on by saying:
Determine for which natural numbers n the polynomial Pn(x)=(x^n)+(x^n-1)+....x+1 has integer roots and find them. (By an integer root we mean z in Z such that f(z)=0). Give two solutions for the problem
1) By proving a suitable generalization of the already said theorem above.
2) By using complex numbers and the polynomial x^(n+1)-1

Yeah I have no idea what to do here. Any help appreciated.
• Sep 22nd 2006, 11:12 AM
topsquark
Quote:

Originally Posted by JaysFan31

If r is in Z and r is a non-zero solution of x^2+ax+b=0 (where a,b are in Z) prove that r divides b.

I solved it the following way:
Letting c and r be the roots of the polynomial.
(x-c)(x-r)=0
x^2+x(-r-c)+cr=0
cr=b, which is the definition of r dividing b

The question then goes on by saying:
Determine for which natural numbers n the polynomial Pn(x)=(x^n)+(x^n-1)+....x+1 has integer roots and find them. (By an integer root we mean z in Z such that f(z)=0). Give two solutions for the problem
1) By proving a suitable generalization of the already said theorem above.
2) By using complex numbers and the polynomial x^(n+1)-1

Yeah I have no idea what to do here. Any help appreciated.

1) Generalize your proof of the first theorem to show that if r is a non-zero solution of x^n + ax^{n-1} + ... + b = 0 that r divides b. (r,a,...,b in Z).

Now, given the polynomial equation x^n + x^{n-1} + ... + 1 = 0 we know that for any non-zero r that solves the equation, r must divide 1. (And r is an integer.) Which integers do this? What then can you say about n?

-Dan
• Sep 22nd 2006, 12:15 PM
ThePerfectHacker
Quote:

Originally Posted by JaysFan31

The question then goes on by saying:
Determine for which natural numbers n the polynomial Pn(x)=(x^n)+(x^n-1)+....x+1 has integer roots and find them.

You can factor,
x^{n+1)-1=0 as,
(x-1)(x^n+x^{n-1}+...+x+1)=0
Now,
The above equation can be solved by de Moiver's theorem.
It has two intergral solutions x=1,x=-1 for n being even and 1, x=1 for n being odd. Thus, since x=1 is already a solution, thus the other is -1 if it exists and n+1 is even. Thus,
x^n+x^{n-1}+...+x+1=0
Has only other integral solution when n=>1 is odd and that solution is x=-1.
Note, it cannot have any other solutions because it is impossible by de Moiver's theorem.
• Sep 22nd 2006, 12:18 PM
ThePerfectHacker
Quote:

Originally Posted by JaysFan31

If r is in Z and r is a non-zero solution of x^2+ax+b=0 (where a,b are in Z) prove that r divides b.

I solved it the following way:
Letting c and r be the roots of the polynomial.
(x-c)(x-r)=0
x^2+x(-r-c)+cr=0
cr=b, which is the definition of r dividing b

I do not favor such an approach because you never did show that it has roots :eek: in the integral domain Z.
(Though you can work on the proof so that it can become acceptable).

There is an eaiser way,
by definition the evaluation homomorphism is zero thus,
r^2+ar+b=0
Thus,
r(-r-a)=b
We say that r divides b because there exists an integer,
-r-b
such that,
r(-r-a)=b
• Sep 22nd 2006, 12:30 PM
topsquark
Quote:

Originally Posted by ThePerfectHacker
I do not favor such an approach because you never did show that it has roots :eek: in the integral domain Z.

Actually (x - c)(x - r) = x^2 + ax + b implies that c is an integer by the division algorithm. It boils down to the same condition that you gave: r(-r-a)=b. (Or were you objecting that this was the part of the argument that was missing?)

-Dan
• Sep 22nd 2006, 12:39 PM
ThePerfectHacker
Quote:

Originally Posted by topsquark
(Or were you objecting that this was the part of the argument that was missing?)

There is nothing wrong with what he said it just does not look like the best way to approach the problem.