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Problem

Find the greatest common divisor of the numbers: $2^{2^{2019}}-1, 2^{2^{2021}}-4$.

My Idea

To be honest, I didn't know how to start so I let $d$ be the greatest common divisor of those numbers

This means that $d| 2^{2^{2019}}-1$ and I multiply by $2^{2^{2019}*3}$ and got that $d| 2^{2^{2021}}- 2^{2^{2019}*3}$

I also know that $d| 2^{2^{2021}}-4$ and I thought of decreasing the results I got and getting that

$d| 2^{2^{2019}*3} -4$

One thing I observed is that $2^{2^{2019}}-1$ is odd, while $2^{2^{2021}}-4$ is even, I don't know if this might help.

I hope one of you can help me! Thanks!

Bill Dubuque
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IONELA BUCIU
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  • $(2^k!-!1,2^{4k}!-!4) = (2^k!-!1,2^{4k-2}!-!1) = 2^{(k,2)}!-!1,$ by linked dupe and $(k,4k!-!2)=(k,2)\ \ $ – Bill Dubuque Jul 18 '24 at 16:19
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    I am voting to reopen , because that Dup target is quite far from the question here which is "Sufficiently Different". We can , of course , solve all general gcd questions using general tools. Particular questions may allow Specific Solutions like here. I do not think the Solution given here will fit that Dup target. In my opinion , this is not a Dup ! – Prem Jul 18 '24 at 16:38
  • Or by gcd mod reduction $\ ((\color{#0a0}{2^k})^4!-!4,:!\color{#0a0}{2^k}!-!{\color{#c00}{\bf 1}}) = (\color{#c00}{\bf 1}^4!-!4,:!\color{#0af}2^k!-!1) = (3,:!(\color{#0ad}{-1})^k!-!1)\ [= 3,\ {\rm by},\ 2\mid k]$ – Bill Dubuque Jul 18 '24 at 16:58
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    My hint would be: if you let $x := 2^{2^{2019}}$, then $x \equiv 1 \pmod{d}$ and also $x^4 = 2^{2^{2021}} \equiv 4 \pmod{d}$. – Daniel Schepler Jul 18 '24 at 17:26
  • @Dan that's the first gcd mod reduction step in my prior comment, i.e. $$\begin{align} \bmod \color{#0a0}{2^k}!-!\color{#c00}1!:\ \ \color{#0a0}{2^k}\equiv \color{#c00}1&,\Rightarrow, (\color{#0a0}{2^k})^4\equiv\color{#c00}1^4\[.4em]
    {\rm so}\ \ (\underbrace{(\color{#0a0}{2^k})^4}_{\large \color{#c00}1^4}!-!4,:!\color{#0a0}{2^k}!-!{\color{#c00}{\bf 1}}) &= (\color{#c00}{\bf 1}^4!-!4,:!2^k!-!1)\end{align}\qquad\qquad$$     – Bill Dubuque Jul 18 '24 at 17:35

1 Answers1

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Let $X = 2^{2^{2019}}-1$ , then $X+1 = 2^{2^{2019}}$

Let $Y = 2^{2^{2021}}-4$ , then $Y = 2^{2^{2021}}-4 = 2^{2^{2019}\times2^2}-4 = (2^{2^{2019}})^{2^2}-4$

$Y = (X+1)^{2^2}-4 = (X+1)^{4}-4$

$Y = (X^4+4X^3+6X^2+4X+1)-4$
$Y = X^4+4X^3+6X^2+4X-3$
$3 = X^4+4X^3+6X^2+4X - Y$

When $X$ & $Y$ have some factor in common , then $3$ too must have that factor in common.

We just have to check whether $3|X$ & $3|Q$ , then $gcd = 3$ , else $gcd = 1$

Prem
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  • Thanks for your answer! Can you explain this part ,,When $X$ & $Y$ have some factor in common , then $3$ too must have that factor in common."? – IONELA BUCIU Jul 18 '24 at 13:03
  • $3 = X^4+4X^3+6X^2+4X - Y$ indicates that $3$ occurs on either side. – Prem Jul 18 '24 at 13:05
  • Yeah, but how this does this mean that the factor common that X and Y have is one 3 must also have...? – IONELA BUCIU Jul 18 '24 at 13:09
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    Can $X$ & $Y$ have common factor $5$ or $7$ or $11$ or more ? Will such common factor allow $3 = X^4+4X^3+6X^2+4X - Y$ , where $3$ is on one side yet other side has higher factor ? WIll it occur that $3 = 5Z$ or $3 = 7^4+7^3 \cdots -7$ or $3 = 11U-11V$ ? – Prem Jul 18 '24 at 14:38
  • Please strive not to post more (dupe) answers to dupes of FAQs. This is enforced site policy, see here. It's best for site health to delete this answer (which also minimizes community time wasted on dupe processing.) – Bill Dubuque Jul 18 '24 at 16:20
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    I get what you imply , though that Dup is quite far from the question here which is "Sufficiently Different". We can , of course , solve all general gcd questions using general tools. Particular questions may allow Specific Solutions like here. I do not think the Solution given here will fit that Dup , @BillDubuque , In other words , this is not a Dup. – Prem Jul 18 '24 at 16:32
  • See the closing comment - it follows immediately from the dupe. Please don't destroy site organization by voting to reopen dupes. – Bill Dubuque Jul 18 '24 at 16:40
  • @BillDubuque May i ask how i can check that X and Y are divisible by 3? Im not sure if my idea is right: $2$ is congruent with $-1$ modulo 3 which makes $x+1$ congruent with 1 modulo 3 so x divisible with 3. We do the same thing for y. Is my idea right? – IONELA BUCIU Jul 18 '24 at 17:10
  • @BillDubuque But still, is my idea of showing that X and Y are divisible by 3? Also, is his idea right?\ – IONELA BUCIU Jul 18 '24 at 17:13
  • @BillDubuque Let me explain: I followed his solution till the point where he said that ,,When X & Y have some factor in common , then 3 too must have that factor in common." This means that the gcd is either 3 if we can show that X and Y are divisible by 3, either 1 if we don't show this. That's why I asked if my idea of showing that X and Y are divisible is right. Is his idea right? – IONELA BUCIU Jul 18 '24 at 17:20
  • @ION This answer can be viewed more clearly as applying the Euclidean algorithm, i.e. applying $$\begin{align} &!!!(x,,f(x)) ,=, (x,, f(x)\bmod x) ,=, (x,f(0))\[.5em] {\rm so\ for}\ \ y = f(x)&=(x!+!1)^4-4\ \text{ we get }\ (x,y)=(x,f(0)) = (x,3)\end{align}\qquad$$ This is exactly the same gcd mod reduction method I used in my 2nd comment on the question, i.e. $(x,y) = (x,\bar y)\ $ if $\ y\equiv \bar y\pmod{!x}.,$ Iterating this yields the Euclidean gcd algorithm when $,\bar y = (y\bmod x).,$ I strongly recommend that you master these general (vs. ad hoc) methods. – Bill Dubuque Jul 18 '24 at 23:30