Give the remainder when $x^{203}-1$ is divided by $x^4-1$.

Question

Give the remainder when $x^{203}-1$ is divided by $x^4-1$.

Now it's easy to see that $P(1)=1^{203}-1=0.$

Writing $P(x)=(x^4-1)Q(x)+R(x)$ I get that $\deg(R(x))$ is at most $3$.

So $R(x)=x^3+ax^2+bx+c$, however I'm not sure what to do with this... How can I use this since now I've just introduced $a,b$ and $c$ also which I cannot seem to find to determine the remainder. Any hints would be appreciated.

See this duplicate for the standard approach. – Dietrich Burde Sep 28 '20 at 08:55 — Dietrich Burde, Sep 28 '20 at 08:55
You can plug in $x=\pm1,\pm i$. – Lynnx Sep 28 '20 at 08:55 — Lynnx, Sep 28 '20 at 08:55
Hint: Start with $$x^4\equiv 1\pmod{x^4-1}$$ – Thomas Andrews Sep 28 '20 at 08:58 — Thomas Andrews, Sep 28 '20 at 08:58

score 4 · Accepted Answer · answered Sep 28 '20 at 08:58

4

Use the roots of $x^4-1$.

Let $x^{203}-1 = P(x)(x^4-1) + Q(x)$ as polynomials. Note that we can substitute complex values and retain equality.

Now, $x^4-1$ has the roots $\pm 1, \pm i$. Substitute each of these in to get: $$ Q(1) = 0 , Q(-1) =2 , Q(i) = -i-1, Q(-i) = i-1 $$

Note that $Q(x)$ is a multiple of $x-1$, so take it as $Q(x) = (ax^2+bx+c)(x-1)$. Now use the other three conditions to find $a,b,c$ and conclude.

Another way to do it, is to do : $$ x^{203}-1 = (x^{203} - x^3) + (x^3-1) $$

and then note that each of $\pm 1, \pm i$ satisfies $x^{203}-x^3 = 0$, so this polynomial is a multiple of $x^4-1$. Hence the remainder is $x^3-1$.

answered Sep 28 '20 at 08:58

Sarvesh Ravichandran Iyer

76,105

Are you denoting $Q(x)$ as the remainder here? – Sep 28 '20 at 09:23
1

@Daniel Yes, sorry for the confusion. I understand that usually $Q$ stands for quotient, but in this case I took $P$ as quotient so $Q$ is the remainder. – Sarvesh Ravichandran Iyer Sep 28 '20 at 09:25

Peter Szilas · Answer 2 · 2020-09-29T07:37:46.760

3

$x^{203}=(x^4)^{50}x^3;$

$x^{203}-1=$

$((x^4-1)+1))^{50} x^3-1;$

All terms except the last term in the binomial expansion of

$((x^4-1)+1)^{50}$

have a factor $x^4-1.$

The remainder is

$1^{50}x^3-1.$

edited Sep 29 '20 at 07:37

answered Sep 28 '20 at 09:26

Peter Szilas

20,344
2
17
28

score 1 · Answer 3 · answered Sep 28 '20 at 08:57

1

$$x^{203}-1=x^{203}-x^3+x^3-1\equiv x^3-1$$

answered Sep 28 '20 at 08:57

Michael Rozenberg

194,933

score 0 · Answer 4 · answered Sep 28 '20 at 09:35

0

HINT: $$\frac{x^{203}-1}{x^4-1}=\frac{x^{202}+x^{201}+x^{200}+......1}{x^3+x^2+x+1}$$

answered Sep 28 '20 at 09:35

E.H.E

23,280

Give the remainder when $x^{203}-1$ is divided by $x^4-1$.

4 Answers4