# Show that the only divisors are $1$ and $2$ for $(z^{(2^x)}+1)$ and $(z^{(2^y)}+1)$ where $x,y,z\in\mathbb{N}$

I am trying to show that the only divisors are $$1$$ and $$2$$ for both $$(z^{(2^x)}+1)$$ and $$(z^{(2^y)}+1)$$ where $$x,y,z\in\mathbb{N}$$. To start the problem, the logical choice is to use difference of squares. We see that $$z^{2^x}+1-(z^{2^y}+1)=z^{2^x}-z^{2^y}$$. I am not sure where to go from here except to show that the gcd is 2, which proves the result.

Let $$x Let $$n=\gcd(1+z^{2^x},1+2^{2^y}).$$ Then $$z^{2^x}\equiv -1 \equiv z^{2^y} \pmod n.$$ Now $$y-x-1$$ is a non-negative integer, so modulo $$n$$ we have $$-1\equiv z^{2^y}\equiv (\,(z^{2^x})^{2^{y-x-1}}\,)^2\equiv$$ $$\equiv (\,(-1)^{2^{y-x-1}}\,)^2\equiv$$ $$\equiv (\,\pm 1\,)^2 \equiv 1.$$
So $$-1\equiv 1 \pmod n.$$ And since $$d\equiv e \pmod n \iff n|(e-d)$$ (for any $$d,e$$), we have $$n|(1-(-1))=2,$$ so $$n\le 2.$$
We have used the fact that for any $$a,b, n\in \Bbb Z$$ with $$n\ne 0,$$ and any $$c\in \Bbb N,$$ if $$a\equiv b \pmod n$$ then $$a^c\equiv b^c \pmod n.$$
• This is also valid, verbatim, if $0=x<y.$ – DanielWainfleet Mar 14 at 6:48