Remainder of a Polynomial of a very high degree on division

Question

The question is find the remainder when $x^{60}-1$ is divided by $x^3-2$. My intuition is that this expression can be written as $a^n-\frac{b^m}{a}-b$. However, I tried a bunch of ways and I haven't been able to factor the numerator into that form. This is from an olympiad problem book on Higher Algebra.

Answer 1

$$x^{60}=(2+x^3-2)^{20}\equiv2^{20}\pmod{x^3-2}$$

Alternatively, using Why is $a^n - b^n$ divisible by $a-b$?,

$$(x^3)^{20}-2^{20}$$ is divisible by $x^3-2$

Answer 2

We can use the remainder theorem. The outcome of dividing the polynomial $p(x)$ by another polynomial $d(x)$ is essentially an equation $$p(x)=q(x)d(x)+r(x)$$where the degree of $r(x)$ is less than the degree of $d(x)$.

Suppose $d(a)=0$. It follows that $p(a)=r(a)$. So one can reconstruct $r(x)$ by evaluating $p(x)$ at the roots of $d(x)=0$ (with an additional twist if there are multiple roots).

Here, setting $y=x^3$ we can simplify considerably by using $p(y)=y^{20}-1$, $d(y)=y-2$ and then degree $(r)\lt 1$ so $r$ must be a scalar.

We obtain $r=p(2)=2^{20}-1$

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I mentioned reconstructing a remainder which is not a scalar above, because not everyone realises the remainder theorem can be used in this way, and these kinds of remainder questions have occurred in competition problem sets.

Answer 3

It is the same as if we ask what is the remainder if $t^{20}-1$ is divided by $t-2$ and $t=x^3$. So by the remainder theorem we have: $$t^{20}-1= k(t)(t-2)+c$$ where $c$ is constant. Pluging $t=2$ we get $c= 2^{20}-1$.