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Suppose that $d_n = \gcd(3\cdot2^n + 1, 2^{2n}- 3)$, where $n>0$.

Show that $d_n$ is either $1$ or $13$.

I tried to use the fact that $\gcd(a,b)=\gcd(a,a-b)$, but I couldn't go much further than that. How can I do this proof?

Anne Bauval
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Diego Cândido
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  • Hint: $9(2^{2n}-3)\equiv-26\bmod{3\cdot2^n + 1}$. – Anne Bauval Apr 11 '24 at 22:08
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    Welcome to Mathematics Stack Exchange. Do you know the Euclidean algorithm? $26=(3\cdot2^n-1)(3\cdot2^n+1)-9(2^{2n}-3)$ – J. W. Tanner Apr 11 '24 at 22:09
  • Mimicking the Euclidean algorithm as here in the linked dupe, we can easily derive that $(3x!+!1,x^2!-!3) = (3x!+!1,26)\ [= (3x!+!1,13),$ here by $,3x!+!1 = 3\cdot 2^n+1,$ odd]$\ \ $ – Bill Dubuque Apr 11 '24 at 22:20
  • In fact, if you extend the calculation a bit more, you can show $(3x+1, x^2-3) = (x+9, 26)$. – Daniel Schepler Apr 11 '24 at 22:51
  • @Dan Yes, that's not needed here, but is done in the linked dupe. Above we can continue $,(3x!+!1,26) = (3(x!+!\color{#c00}{3^{-1}}),26) = (x!+!\color{#c00}9,26),,$ by Euclid and $!\bmod 26!:\ \color{#c00}{\dfrac{1}3\equiv \dfrac{27}3\equiv 9}\ \ $ – Bill Dubuque Apr 12 '24 at 03:41
  • Or $\ d\mid \color{#c00}{x^2!-!3},\color{#0a0}{3x!+!1}\Rightarrow\bmod d!:,\color{#0a0}{-1\equiv 3x}\overset{(\ \ )^{\large 2}}\Rightarrow 1\equiv 9\color{#c00}{x^2}\equiv 9\cdot \color{#c00}3\Rightarrow 26\equiv 0\Rightarrow d\mid 23,,$ which is a shorter way vs. Bezout in a comment @J.W.T As explained here this can be viewed as taking a norm, i.e. $,\color{#0a0}\alpha = 3\color{#c00}{\sqrt 3}+1\equiv 0\Rightarrow N(\alpha) = \alpha\bar \alpha\equiv -26\equiv 0\ \ $ – Bill Dubuque Apr 12 '24 at 04:07
  • @BillDubuque: nice. I think you meant $26$ where you typed $23$ – J. W. Tanner Apr 12 '24 at 04:14
  • The method in the comment @Anne is the same method used in the linked dupe, i.e. $\color{#c00}{x^2!-!3}\equiv 0\equiv \color{#0a0}{3x!+!1}\Rightarrow 0\equiv 9(\color{#c00}{x^2!-!3})\equiv (\color{#0a0}{3x})^2!-27\equiv (\color{#0a0}{-1})^2!-!27\equiv -26\ \ $ – Bill Dubuque Apr 12 '24 at 04:17
  • I guess the point I was sort of driving towards is: that expression makes it easy to determine exactly for which $n$ you have $(3\cdot 2^n + 1, 2^{2n} - 3) = 13$: it happens exactly when $2^n \equiv 4 \pmod{13}$, which turns out to be exactly when $n \equiv 2 \pmod{12}$. – Daniel Schepler Apr 12 '24 at 17:21

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