0
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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 $a^c\equiv b^c \pmod n.$

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.