# Show that $(x^{2^k}+1)\mid (x^{2^l}-1)$, when $k<l$ [duplicate]

I found this in some notes from a course in number theory. How do i work to solve this?

• Write $x^{2^k}=y$. Can you show that $y+1\mid y^2-1\mid y^{2^{\ell-k}}-1$? – Jyrki Lahtonen Apr 17 '19 at 17:15
• Meaning that the question is reduced to this near duplicate. – Jyrki Lahtonen Apr 17 '19 at 17:22
• Also, I highly recommend that you take a look at our guide for new askers. The question is a bit lacking in the context department. If you could convince us that you are not just trying to get somebody to do your homework we would feel a lot better about the question. – Jyrki Lahtonen Apr 17 '19 at 17:24

$$x^{2^l}-1=(x^{2^{l-1}}+1)(x^{2^{l-1}}-1)$$ Continue to expand the right hand factor until you get $$x^{2^l}-1=(x-1)\prod_{k=0}^{l-1} \left(x^{2^k}+1\right)$$ Which is obviously divisible by $$x^{2^k}+1$$ for all $$0\le k\lt l$$.
$$\bmod\, x^{\large 2^{\Large K}}\!\!+1\!:\ \ \color{#c00}{x^{\large 2^{\Large K}}\!\!\equiv -1}\,\Rightarrow\, x^{\large 2^{\Large K+N}}\!\!\equiv (\color{#c00}{x^{\large 2^{\Large K}}})^{\large 2^{\Large N}}\!\!\equiv (\color{#c00}{-1})^{\large 2^{\Large N}}\!\!\equiv 1\$$ $$\!\!\overbrace{{\rm when} \ \ N> 0}^{\large K\, <\, K+N\, =:\, L_{\phantom{I_I}}}$$
Remark  It's a special case of $$\ x^{\large K}\!+1\mid x^{\large 2K}\!-1\mid x^{\large 2KN}\!-1,\,$$ also provable by mod
$$\bmod\, x^{\large K}\!+1\!:\ \ \color{#c00}{x^{\large K} \!\!\equiv -1}\,\Rightarrow\, x^{\large 2KN} \!\equiv (\color{#c00}{x^{\large K}})^{\large 2N}\!\equiv (\color{#c00}{-1})^{\large 2N}\!\equiv 1\$$
• If you learn to reason by $\!\bmod$ as above then you don't have to remember motley divisibility formulas - they occur very naturally as special cases of general results (e.g. abobe that $\,(-1)^{2N} = 1)\ \$ – Gone Apr 17 '19 at 17:40