Proof that a polynomial is a factor

[mindauggas]

Homework Statement

Show that $x+a$ is a factor of $x^{n}+a^{n}$for all odd n.

The Attempt at a Solution

(1) Assume that $x+a$ is a factor of $x^{n}+a^{n}$for all odd n. This implies that when $x^{n}+a^{n}$ is divided by $x+a$ the remainder is zero.

I don't know - is this a sensible 1st step?

What to do next?

 

[genericusrnme]

I'd go for contradiction, suppose it isn't and see what happens

 

[mindauggas]

Could you give me some hint what to do next? I really don't understand how to incorporate the "for all odd n" part into the reasoning process.

 

[genericusrnme]

What property do things with odd powers have that things with even powers don't?

 

[tiny-tim]

hi mindauggas!

hint: roots? ​

 

[HallsofIvy]

As tiny tim suggests, x- a is a factor of polynomial p(x) if and only if p(a)= 0. Of course, x+ a= x-(-a).

 

[mindauggas]

What property do things with odd powers have that things with even powers don't?

Probably a lot of properties (e.g. the "thing" with odd power is not always positive after raising it to the power and the "thing" with even powers is always positive ("thing" in ℝ)).

Presumably you are asking the most general property that distinguishes them ... this I beg you to tell me, please (this would help me not only in this problem but in general).

 

[genericusrnme]

Probably a lot of properties (e.g. the "thing" with odd power is not always positive after raising it to the power and the "thing" with even powers is always positive ("thing" in ℝ)).

Yes, you've got it!
How can you use that to arrive at a contradiction now?

 

[mindauggas]

As tiny tim suggests, x- a is a factor of polynomial p(x) if and only if p(a)= 0. Of course, x+ a= x-(-a).

So I just put - $f(-a)=(-a)^{n}+a^{n}=0$, under the condition that n is odd (now I see how that came into play) and that's it?

Thanks.