Any number prime can only be written as $10k + x$ for $x=1,3,7,9$ since others cannot be primes. Is there a way to prove that there are infinitely many primes of the form $10k+3$ or $10k+7$?
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See also this question. – Dietrich Burde Mar 18 '18 at 15:10
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It follows again from Dirichlet's theorem, see here. – Dietrich Burde Mar 18 '18 at 15:21
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@DietrichBurde I know the theorem but I wanted a proof. Thanks for the tip though – mandella Mar 18 '18 at 15:51
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This comes from Dirichlet's theorem, but there is an elementary way to prove that the set $$A=\{n\in\Bbb N:n\equiv\pm3\pmod{10}\}$$ contains infinitely many primes. Prove that each number of the form $$10N+3$$ has a prime factor $p$ in the set $A$. Now take $N$ to be the product of some primes in $A$ but excluding $3$. Then $p$ will be a further prime ($\ne 3$) in $A$.
Angina Seng
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Yes. Let $F$ be a finite set of primes of the form $10k+3$ or $10k+7$. Let $P$ be the product of the elements of $F\setminus\{3\}$ (which is simply $F$ if $3\notin F$) and let $N=10P+3$. Note that $N$ is even and that $5\nmid N$. On the other hand, not all prime factors of $N$ are of the type $10k+1$ or $10k+9$, because the product of several numbers of this type is again of this type. Therefore, $N$ has a prime factor $p$ which is of one the forms $10k+3$ and $10k+7$. Furthermore, $p\neq3$, because$$3\mid N\iff3\mid10P\iff3\mid P,$$which is impossible, since $P$ is a product of prime numbers, none of which is $3$.
So, $N$ has a prime factor $p$ which is of the form $10k+3$ (but distinct from $3$) or of the form $10k+7$. But then $p\notin F$, because\begin{align}p\in F\iff&p\in F\setminus\{3\}\\\implies&p\mid P\\\implies&p\mid10P\\\implies&p\mid3\text{ (because $3=N-10P$)}\\\implies&p=3,\end{align}which is false.
José Carlos Santos
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I looked more closely, and I cannot follow why $p | P \Rightarrow p | 5$? Is it because it divides $N$? – mandella Mar 18 '18 at 16:03
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1@mandella Yes. Since $p\mid N$ and $p\mid10P$ (because $p\mid P$), $p\mid(N-10P)$, which means that $p\mid5$. – José Carlos Santos Mar 18 '18 at 16:05
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And also, $p \notin F$ follows because if $p | 5 $ then $p$ is either $5$ or $1$, neither of which can be in the set, correct? – mandella Mar 18 '18 at 17:06
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1@mandella Almost. The only possibility is $p=5$, since I assumed that $p$ is prime. But $5\notin F$. – José Carlos Santos Mar 18 '18 at 17:15
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Yes I forgot that it was supposed to be a prime and $5$ does not have the form $10k+3$ or $10k+7$ – mandella Mar 18 '18 at 17:21
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@Wojowu Actually, I've just proved that it can't be a power of $5$. The relevance of this still eludes me, though. – José Carlos Santos Mar 18 '18 at 19:57
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You claim that $N$ has a prime factor of the form $10k+3$ or $10k+7$. However, I don't think this conclusion holds - $N$ may only have prime factors $10k+1,10k+9$ and $5$. – Wojowu Mar 18 '18 at 20:17
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@Wojowu You are right. I shall edit my answer, but I can't do it right now. – José Carlos Santos Mar 18 '18 at 20:53
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@Wojowu what happens in that case? I thought it wasn't in the set at all – mandella Mar 18 '18 at 21:19
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@JoséCarlosSantos could you explain what happens in this case bc I have a quiz tomorrow and would like to know this – mandella Mar 18 '18 at 21:28
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Could you explain why since $5$ is not the only prime factor, it follows that $N$ can not have prime factors of the form $10k + 1$ or $10k + 9$? – mandella Apr 03 '18 at 06:17
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1@mandella No, I cannot. I never claimed that. All I wrote was that, since $5$ is not the only prime factor, then $N$ must have a prime factor of the form $10k+3$ or of the form $10k+7$. – José Carlos Santos Apr 03 '18 at 06:29
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Also another question. If we take $\frac{N}{5}$ why can't it have prime factors only of the form $10n+1$ and $10n+9$? @JoséCarlosSantos – mandella Apr 03 '18 at 13:12
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@mandella Why? Suppose that $F={3,7,13,17}$. Then $N=46,415$ and one of its prime factors is $9,283$. – José Carlos Santos Apr 03 '18 at 13:26
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Yes but then this implies that our number can have such factors, so doesn't need to have prime factors $10k+3$ or $10k+7$ – mandella Apr 03 '18 at 13:26
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@mandella Didn't you notice that $9,283$ is of the form $10k+3$? – José Carlos Santos Apr 03 '18 at 13:28
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I do not see why $\frac{N}{5}$ cannot have ONLY prime factors $10k+1$ or $10k+9$ ... @JoséCarlosSantos – mandella Apr 03 '18 at 16:11
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1@mandella I made a mistake and I will correct it within less than a hour. – José Carlos Santos Apr 03 '18 at 16:12
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1@mandella Done. I believe that it is correct now, but if it isn't, then please unmark my answer as the accepted on, so that I can delete it. And please excuse me for wasting your time. – José Carlos Santos Apr 03 '18 at 16:48
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