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 $x<y.$ 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 $a^c\equiv b^c \pmod n.$

  • $\begingroup$ This is also valid, verbatim, if $ 0=x<y.$ $\endgroup$ – DanielWainfleet Mar 14 at 6:48

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