If $p(x)$ be the polynomial left as remainder when $x^{2019}-1$ is divided by $x^6+1$. What is the remainder left when $p(x)$ is divided by $x-3$?

Question

If $p(x)$ be the polynomial left as remainder when $x^{2019}-1$ is divided by $x^6+1$. What is the remainder left when $p(x)$ is divided by $x-3$?

It is obvious that remainder is a constant. How do I solve it?

Answer 1

The other answers give cooler results, but keeping in mind the algebra-precalculus tag:

$$x^{2019}-1=(x^6+1)(x^{2013}-x^{2007}+\cdots+(-1)^kx^{2013-6k}+\cdots+x^9-x^3)+(x^3-1)$$ $$\therefore x^3-1=(x-3)(x^2+3x+9)+26$$

Answer 2

You can write $x^{2019} - 1$ in the form:

$$x^{2019} - 1 = x^{2013} (x^6 + 1) - x^{2013} - 1$$

and $x^{2013}$ can be written in the form:

$$x^{2013} - 1 = x^{2007} (x^6 + 1) - x^{2007} - 1$$ $$\Rightarrow x^{2019} - 1 = x^{2013} (x^6 + 1) - \left(x^{2007} (x^6 + 1) - x^{2007}\right) - 1$$ $$= (x^6 + 1)(x^{2013} - x^{2007})+ x^{2007}-1$$

Each time this process is applied, you can isolate $x^{n - 12}$ from $x^{n}$, where $n$ is a natural number. Does this help with understanding the explanation in the comments?

After every multiple of $x^{12}$, the sign of the remainder term is unchanged. This justifies the remainder as being $x^3 - 1$ instead of $-(x^3 - 1)$. This is separate from the quotient $x^{2013} - x^{2007} + x^{2001} \cdots$ where the signs do indeed alternate.

Answer 3

Let $t=x^3$, then we need to find a remainder if we divide $t^{673}-1$ with $t^2+1$. Write:

$$t^{673}-1= k(t)(t^2+1)+q(t)$$ where $q$ is first degree polynomial $q(t)=at+b$, where $a,b$ are clearly real. Now plug $t=i$ in equation in order to find $a$ and $b$:

$$ i-1 = k(i)\cdot 0 +ai+b\implies a=1 \;\;\;{\rm and}\;\;\;b=-1$$

Since $p(x) = q(t) = x^3-1$ we are done, $p(3)=26$.

Answer 4

As $12|2016$, then $x^{12}-1|x^{2016}-1$, and because $x^{12}-1=(x^6-1)(x^6+1)$ we have $x^6+1|x^{2016}-1$.

Therefore $x^6+1|x^3(x^{2016}-1)=x^{2019}-x^3$, which gives the remainder when $x^6+1$ divides $x^{2019}-1$ as $x^3-1$.

$(x-3)^3=x^3-9x^2+27x-27$ which implies $0\equiv x^3-27 \pmod{x-3}$ as $-9x^2+27x=-9x(x-3)$.

So $x-3|x^3-27$ and as $x^3-1=x^3-27+26$, and the final remainder is $26$.