High degree polynom divisibility

Question

The problem is:

Show that $x^{44}+x^{33}+x^{22}+x^{11}+1$ is divisible by $x^4+x^3+x^2+x+1$

I am not sure how to approach this problem so any help would be very appreciated. Thank you in advance.

Did you mean $x^{44}+x^{33}+x^{22}+x^{11}+1$? – mathematics2x2life Feb 10 '14 at 05:15 — mathematics2x2life, Feb 10 '14 at 05:15
Yes, I did. Thank you for noticing. – suomynona Feb 10 '14 at 05:30 — suomynona, Feb 10 '14 at 05:30

score 3 · Answer 1 · edited Apr 13 '17 at 12:21

3

HINT:

Both are in Geometric Series

$$1+x+x^2+x^3+x^4=\frac{1-x^5}{1-x}$$ and $$1+x^{11}+x^{22}+x^{33}+x^{44}=\frac{1-x^{55}}{1-x^{11}}$$

So, $$\frac{1+x^{11}+x^{22}+x^{33}+x^{44}}{1+x+x^2+x^3+x^4}=\frac{(1-x^{55})(1-x)}{(1-x^{11})(1-x^5)}$$

Now using Prove that $\gcd(a^n - 1, a^m - 1) = a^{\gcd(n, m)} - 1$,

$1-x^{55}$ is divisible by $1-x^5,1-x^{11}$ and

$(1-x^{11},1-x^5)=1-x$

edited Apr 13 '17 at 12:21

Community

1

answered Feb 10 '14 at 05:22

lab bhattacharjee

274,582

You need more than just that to do it this way, e.g. see Andre's more complete sketch. But it is much easier by modular arithmetic - see my answer. – Bill Dubuque Feb 10 '14 at 05:48
@BillDubuque, please find the edited answer – lab bhattacharjee Feb 10 '14 at 05:58
That's better. It is important to explicitly mention the required coprimality since many students overlook that essential point. – Bill Dubuque Feb 10 '14 at 06:06

score 2 · Answer 2 · answered Feb 10 '14 at 05:38

2

${\rm mod} \dfrac{(x^5\!-\!1)}{(x\!-\!1)}:\,\ \color{#c00}{x^5\equiv 1}\,\Rightarrow\,x^{11n}\equiv x^n (\color{#c00}{x^5})^{2n}\equiv x^n\Rightarrow\,x^{44}\!+\!x^{33}\!+\cdots+1\equiv x^4\!+\!x^3\!+\cdots+1\equiv 0$

answered Feb 10 '14 at 05:38

Bill Dubuque

272,048

Thank you, this really is a rather clear and nice answer. – suomynona Feb 10 '14 at 19:08

score 1 · Answer 3 · answered Feb 10 '14 at 05:38

It is not hard to see that $x^5-1$ divides $x^{55}-1$.

Thus $(x-1)(x^4+x^3+x^2+x+1)$ divides $(x^{11}-1)(x^{44}+x^{33}+x^{22}+x^{11}+1)$ and therefore

$x^4+x^3+x^2+x+1$ divides $(x^{10}+x^9+\cdots +x+1)( x^{44}+x^{33}+x^{22}+x^{11}+1)$.

But $x^4+x^3+x^2+x+1$ and $x^{10}+x^9+\cdots +x+1$ are relatively prime (use the Euclidean Algorithm).

The result follows.

score 0 · Answer 4 · answered Feb 10 '14 at 05:19

0

Try plain polynomial long division. If it gives you trouble, try it with a smaller example like $x^4-2x^3+4x^2-x+7$ divided by $x^2-3$.

In your original problem, you should find the quotient to be....

$x^{40}-x^{39}+x^{35}-x^{34}+x^{30}-x^{28}+x^{25}-x^{23}+x^{20}-x^{17}+x^{15}-x^{12}+x^{10}-x^6+x^5-x+1$

answered Feb 10 '14 at 05:19

mathematics2x2life

13,322

I thought about this, but wondered if there was a more "elegant" way to do it. Thank you anyway though. – suomynona Feb 10 '14 at 05:28
Indeed, this is an easy way to do it. Given the polynomials are simple, the division goes quickly. However, lab bhattacharjee's answer may provide a shorter way. It's certainly true and may indeed yield a much faster answer! However, I have yet to see how it gets you there faster. But it's certainly worth a strong look! – mathematics2x2life Feb 10 '14 at 05:34
1

There is a very short way: compute the remainder only (we don't need the quotient) by using modular arithmetic. This yields a simple one-line proof - see my answer. @suomynona too. – Bill Dubuque Feb 10 '14 at 06:27

High degree polynom divisibility

4 Answers4