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I am asked to show that $42\mid (d_{21092020} +1)$, for $d_m = 5+ 2\cdot 3^m$, here is a proof I have to show this, using modular arithmetic. I want to make sure my reasoning is sound, and correct any errors in my logic.

$42\mid (d_{m} +1)$

$\iff$ $\bar{d_m}+\bar{1} = \bar{0}$ . Where $\bar{}$ denotes $mod42$. Hence by definition of $d_m$, we have that

$\bar{5}+\bar{2}\cdot \bar{3}^m +\bar{1} = \bar{0}$

$\Rightarrow$ $\bar{2}\cdot \bar{3}^m = -\bar{6}$ $\implies$ $\bar{3}^m = -\bar{3}$.

$\bar{3}^m$ is cyclic with period 6, and that for $m=6k+4$ for $k\in \mathbb{Z}$, then the relationship $\bar{3}^m = -\bar{3}$ holds. Notice that we can write $21092020=6k+4$, with $k=3515336$. Thus since $21092020$ can be written in this manner, then it must be true that $42\mid (d_{21092020} +1)$.

Is this enough for a relatively basic proof? Do I need to add anything more?

icarus
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  • It is not true that $3^m\equiv -3\times 3^m\pmod {42}$. – lulu Sep 29 '21 at 00:14
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    Perhaps you would find it simpler to note that $42=2\times 3\times 7$. Now it is obvious that $2$ and $3$ divide your expression, so you just have to check $7$. – lulu Sep 29 '21 at 00:15
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    Hint: $42 | 6n$ iff $7 | n$. – eyeballfrog Sep 29 '21 at 00:15
  • Hint: $a^{p-1} \equiv 1 \mod p$ – Derek Luna Sep 29 '21 at 00:19
  • @lulu That could also definitely work, thanks for pointing it out! – icarus Sep 29 '21 at 00:29
  • And there was a period at the end of $\bar{3}^m = -\bar{3}$, I wasn't saying $\bar{3}^m = -\bar{3} \cdot \bar{3}^m$, made it clear now hopefully. – icarus Sep 29 '21 at 00:29
  • Dividing by $2$ in the last implication is not valid because $2$ is not coprime to $42$. – Geoffrey Trang Sep 29 '21 at 00:36
  • It's clearly divisible by $6$ so you need only check divisibility by $7,,$ whuch follows immediately by mod order reduction in the first linked dupe, i.e $,2^{\color{#c00}{6}}\equiv 1\pmod 7,$ so $,n\equiv 4\pmod 6\Rightarrow 3^n\equiv 3^4\equiv 4,$ so $,2\cdot 3^n+6\equiv 2\cdot 4+6\equiv 0\ \ $ – Bill Dubuque Jan 09 '22 at 16:49

2 Answers2

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A much simpler way that you would establish that $42$ divides $N \doteq 2×3^{21092020}+6$ is to show that both $6$ divides $N$, and $7$ divides $N$.

Now $6$ divides $N$ as you can easily see. The crux is now showing that $7$ divides $N$, so this is what we show next. Now $$N \equiv_7 2×3^{21092020} +6 $$ so let's evaluate $3^{21092020} \pmod 7$. To this end, note that $$3^{21092020} \equiv_7 3^{21092020 \pmod 6}, $$ and so as $21092020 \equiv_6 4$, it follows that $$3^{21092020} \equiv_7 3^4 =81 \equiv_7 4,$$, and so $$3^{21092020} \pmod 7 = 4.$$ Thus, $$N \equiv_7 2×(3^{21092020} \pmod 7) +6$$ $$\equiv_7 (2×4) + 6 \equiv_7 1-1 =0.$$ So $N \equiv_7 0$ and thus $7$ indeed divides $N$. And, as both $6$ and $7$ divide $N$, it follows that $42$ indeed divides $N$ as well.

Mike
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let $n=2*3^m+6$
$m\equiv 4\pmod{6}$
by FLT, $3^m\equiv 3^4\equiv 4\pmod{7}$
$n\equiv 2*4+6\equiv 0\pmod{7}$
since n is divisible by 2,3,7, it is divisible by 42.

cineel
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