A Number Theory problem (GCD)

Question

Prove that $$gcd(2^{2^m}+1, 2^{2^n}+1)=1$$ if $m, n$ are positive integers such that $m \neq n$.

A Hints to solve the problem is also given in the book as follows:

Let $m>n$. Then $2^m=2^{n}2^{m-n}=2^n.2k$ for some integer $k$, (since $m>n$). Let $2^{2^n}=x$. Then $2^{2^m}-1=x^{2k}-1$ and it is divisible by $x+1$.

I am still unable to solve. Please help.

You better add the condition $m\ne n$ since the statement is trivially false for $m=n$. — Rory Daulton, Mar 21 '15 at 15:23
WLOG, assume that $n>m$, then $\gcd(2^{2^n}+1,2^{2^m}+1)=\gcd(2^{2^m}+1,2)=1$. — barak manos, Mar 21 '15 at 15:29
It seems that Fermat has woken up from the dead in order to mark a question concerning his own numbers as duplicate. — barak manos, Mar 21 '15 at 15:42

score 1 · Answer 1 · answered Mar 21 '15 at 15:47

1

Let $A_n = 2^{2^n}+1$. Then: $$ A_{n+h} = (A_n-1)^{2^h}+1,$$ hence for any $h\geq 1$: $$ \gcd(A_{n+h},A_n)=\gcd((-1)^{2^h}+1,A_n)=\gcd(2,A_n)=1$$ since $A_n$ is odd.

answered Mar 21 '15 at 15:47

Jack D'Aurizio

353,855

Please give me the steps to solve according to the Hints given in the book which I have mentioned in my attempt. Thanks all for the above support. – user1942348 Mar 21 '15 at 18:31

score 0 · Answer 2 · edited Apr 13 '17 at 12:20

Lemma $\rm\ \gcd(c+1,\ c^{2J}+1)\ =\ gcd(c+1,\:\color{#b0f}2)$

Proof $\rm\quad\! mod\ c+1\!:\ c^{2J}+1\: \equiv\ (-1)^{2J}+1\:\equiv\ \color{#b0f}2\ $ by $\rm\ c\equiv -1$

For $\rm\,\ \color{#c00}{c=2^{\large 2^{N}}}\!,\ \, 2J\, =\, 2^{K} \Rightarrow\ \color{#0a0}{c^{\large 2J}} = (2^{\large 2^{N}})^{\large 2^K}\!\! = \color{#0a0}{2^{\large 2^{N+K}}}\! $ $\Rightarrow$ $\smash[t]{\, \rm gcd\overset{\underbrace{\large \color{#c00}{c+1}}}{({\it F}_N},\,\overset{\underbrace{\large \color{#0a0}{c^{\Large 2J}+1}}}{{\it F}_{N+K}}) = gcd\overset{\underbrace{\large \color{#c00}{c+1}}}{({\it F}_N},\color{#b0f}2) = 1}$

Remark $\ $ Aternatively, we can employ that $\rm\:c^{2J}+1\: =\: (c^{2J}-1) + 2\:\equiv\: 2\pmod{c+1}\ $
by $\rm\ c+1\ |\ c^2-1\ |\ c^{2J}-1.\,$ But this requires some ingenuity, whereas the above proof does not, being just a trivial congruence calculation using the modular reduction property of the $\rm\,gcd,\:$ namely $\rm\, \gcd(a,b) = \gcd(a,\:b\ mod\ a),\,$ a reduction which applies much more generally. Said equivalently, the result follows immediately by applying a single step of the Euclidean algorithm. Notice how abstracting the problem a little served to greatly elucidate the innate structure.

More generally $\,\rm \gcd(c\!-\!n, f(c))\, =\, \gcd(c\!-\!n, f(n))\ $ for any polynomial $\rm\,f(x)\,$ with integer coeff's, since $\rm\ mod\,\ c\!-\!n\!:\ c\equiv n\,\Rightarrow\,f(c)\equiv f(n)\,$ by the Polynomial Congruence Rule. Above is simply the special case $\rm\, n = -1\,$ and $\rm\, f(x) = x^{2J}\!+1.$

A Number Theory problem (GCD)

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