Let $p_1>p_2>...>p_{n+1}>3$ prime numbers. Show that: $$p_{n+1}\nmid2^{p_1p_2...p_n}+1$$
I tried to prove it by contradiction using Fermat's Theorem but i'm struggling.
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What do you know about the order of $2$ modulo $p_{n+1}$? – Daniel Fischer May 09 '20 at 19:13
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Do you mean $3<p_1<p_2<\dots< p_{n+1}$ ? – snowAuoue May 09 '20 at 19:23
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@DanielFischer it divides $p_{n+1}-1$ – PortoKranto May 09 '20 at 19:26
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@snowAuoue i dont think so – PortoKranto May 09 '20 at 19:26
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That's a good start. So if we had $p_{n+1} \mid 2^{p_1 p_2 \ldots p_n} + 1$, what else could you deduce about the order? – Daniel Fischer May 09 '20 at 19:27
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@DanielFischer it divides $2p_1p_2...p_n$ ?? – PortoKranto May 09 '20 at 19:36
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Correct. And hence it divides … what? – Daniel Fischer May 09 '20 at 19:37
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@DanielFischer so i should say that the order is equal to $2$ because it divides the gcd of the two numbers which is impossible, isn't it ? – PortoKranto May 09 '20 at 19:39
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@DanielFischer or 1 and it is impossible too – PortoKranto May 09 '20 at 19:39
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2Right. It divides the gcd of the two, which is $2$. So the order is either $1$ or $2$, but $2^1-1 = 1$ has no prime factor at all, and $2^2-1 = 3$, however $p_{n+1} > 3$ was assumed. – Daniel Fischer May 09 '20 at 19:42
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@DanielFischer just a question please. It is not obligatory to have $p_1>p_2>..>p_n+1$. They just need to be different and greater and $3$. No ? – PortoKranto May 09 '20 at 19:44
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You need a condition that ensures $p_k \nmid (p_{n+1} - 1)$ for $1 \leqslant k \leqslant n$. If the condition had been stated in that form, the solution would have been obvious, wouldn't it? So a condition was given that implies the above, but isn't obvious. – Daniel Fischer May 09 '20 at 19:48
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@DanielFischer thank you for the help – PortoKranto May 09 '20 at 19:58
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1I added an answer which gives a generalization like that you asked about. – Bill Dubuque May 09 '20 at 21:50
2 Answers
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assume that $p_{n+1}\mid 2^{p_1 p_2\dots p_n}+1\Rightarrow p_{n+1}\mid 2^{2p_1 p_2\dots p_n}-1$, but $p_{n+1}\mid 2^{p_{n+1}-1}-1$, so
$$p_{n+1}\mid\gcd(2^{2p_1 p_2\dots p_n}-1, 2^{p_{n+1}-1}-1)=2^{\gcd(2p_1 p_2\dots p_n, p_{n+1}-1)}-1$$. since $p_{n+1}$ is smallest between these primes, $p_{n+1}-1$ can't have any common factors with $p_1,p_2,\dots p_n$. so $$\gcd(2p_1 p_2\dots p_n, p_{n+1}-1)=2$$ so $p_{n+1}\mid 2^2-1=3$ which is impossible. So we get a contradiction.
snowAuoue
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@N.S. No, $p_{n+1}-1 < p_k$ for any $1\leq k \leq n$, so it can't be divisible by $p_k$. – snowAuoue May 09 '20 at 19:59
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1Tip: when you repeat a proof given in comments it is customary to give credit and make the answer CW. – Bill Dubuque May 09 '20 at 21:53
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Below put $\,p= p_{n+1},\ k = p_1\cdots p_n,\ a = 2$
Lemma $\ $ Prime $\,p\mid a^k+1\,\Rightarrow\, p\mid a^2-1\ $ if $\ p\nmid a\,$ and $\,\color{#c00}{(k,p\!-\!1)=1}.\ $ Proof:
$\!\bmod p\!:\ (a^2)^k\equiv 1\equiv (a^2)^{p-1}\!\Rightarrow\, a^2\equiv 1\,$ by $\,a^2\,$ has order $\,j\!=\!1$, by $\,j\mid \color{#c00}{k,p\!-\!1\ \rm{coprimes}}$
Bill Dubuque
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We used little Fermat and the Order Theorem: $,c^m \equiv 1,\Rightarrow, {\rm ord}(c)\mid m,,$ which also yields a slick proof of the gcd result invoked in the arguments in the comments (repeated in the other answer). – Bill Dubuque May 09 '20 at 21:54