Suppose $n$ is a positive integer. Decide the remainder of $x^{2n}-x^{n}$ when divided by $2$

Question

So, I tried solving this problem in the following way (though I think I'm begging the question...)

The possible remainders of $x^{2n}-x^{n}$ when divided by $2$ is $0$ and $1$. We wish to solve the following congruence $x^{2n}-x^{n}\equiv b\pmod 2$, where $b=0$ or $b=1$.

Now, Suppose $b=0$ (this is where I think I'm begging the question!). Then $x^{2n}-x^{n}\equiv 0\pmod 2 \Leftrightarrow 2\vert (x^{2n}-x^{n}-0) \Leftrightarrow x^{2n}\equiv (x^{n})^{2}\equiv x^{n}\pmod 2$. If we let $a=x^{n}$. Then it is clear that $(x^{n})^{2}\equiv x^{n}\pmod 2 \Leftrightarrow a^{2}\equiv a\pmod 2$ and by Fermat's little theorem we know that $a^{2}\equiv a\pmod 2$ is true. Thus $x^{2n}-x^{n}$ has a remainder of $0$ when divided by $2$.

Is this correct? Or am I just begging the question?

Answer 1

You are overthinking it; just note that

• for a fixed $x$, any power of it has the same parity as $x$
• the difference of two numbers of the same parity is even

Thus, immediately, we see that the remainder is always $0$.

Answer 2

As you said, $x\equiv0$ or $1\pmod2$.

You wish to find $x^{2n}-x^n\pmod2$.

If $x\equiv0\pmod2$, what is $x^n\pmod2$? $x^{2n}\pmod2$? $x^{2n}-x^n\pmod2$?

If $x\equiv1\pmod2$, what is $x^n\pmod2$? $x^{2n}\pmod2$? $x^{2n}-x^n\pmod2$?

Answer 3

$x^{2n}-x^n=x^n(x^n-1)\equiv0\pmod2$.