I was wondering how Euclid showed that there are infinitely many primes by generating a prime number from finitely many primes, and if it could be used to answer if there are infinitely many pairs of primes whose difference is 2. I show my approach in my - short - article  (I know I should copy here the relevant bits, but the article is really short). My question: why is it not proved like this already? Am I missing something?  Update: It seems I was too lazy to check for counterexamples. I removed my article. Thank you for the lot of feedbacks.
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what's the wrong when we use Euclid logic to prove the twin prime conjecture ? $$Q1=p1*p2*p3*.....*pn+1$$ $$Q2=p1*p2*p3*.....*pn-1$$ where: $$Q1-Q2=2$$  Euclid's proof considers any finite set S of primes
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I have heard somewhere that for all primes $p$, for all $k$, there exist $x, y$ s.t. $x^2 + y^2\equiv k \pmod{p}$? I recall that the proof is very elementary, but I can't remember such a proof. How would one prove this? One way is to use Cauchy-Davenport, but I don't think that this is the simplest solution.
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I have seen the past threads but I think I have another proof, though am not entirely convinced.  Suppose there are only finitely many primes $p_1, ..., p_n$ of the form $6k-1$ and then consider the number  $N = (p_1.p_2. ... .p_n)^2 - 1$  If all of the primes dividing N are of the form $6k+1$ then $N$ leaves remainder $1$ upon division by $6$. However, the RHS has remainder $1-1 = 0$ or $-1-1 = -2$ upon division by $6$, depending on whether $n$ is even or odd.  Therefore there must be at least one prime of the form $6k-1$ dividing $N$, and thus some $p_j$ divides N, and so divides $(p_1.p_2. ... .p_n)^2 - N$ which is $1$, a contradiction.  Apologies for this question if it appears unnecessary I'm just beginning number theory and trying to get to grips with it.
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I have shown that the infinite product $$\prod_{p \in \mathcal{P}}\frac{p^2+1}{p^2-1}$$ is equal to $\frac{5}{2}$ (pretty remarkable!). I have checked this numerically with Wolfram Alpha for up to $500000$ primes and it seems true.  I was wondering if this result is recorded anywhere?  Also if true, does this mean that there aren't infinitely many primes of form $p^2+1$?
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Question: Assume you choose any $48$ positive integers so that the number of different prime factors of all those numbers is $10$. Prove or disprove there exists a set of four numbers out of the $48$ positive integers of which product is a perfect square.   I think that it uses pigeon hole, but I couldn't make a progress.
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I'm taking a short number theory course this summer. The first topic we covered was . I still have no sense of why this is important, how it arises, how it might have been discovered, etc. The proof we studied is a bit on the clever side for my taste, giving me no sense of how said proof might have been discovered. Can anyone help?
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Everyone knows that there is an infinitude of primes. I know the Euclide, the Euler and the Erdos proofs. But are they the only known proofs ? I will try, here, to present a fourth one : Let the Fermat numbers $F_n=2^{2^n}+1$. They are not all primes, $F_5$ for example is not, Euler has proved it ! There are two possibilities, of course : if there is an infinitude of prime Fermat numbers, the primes are infinite. If there is an infinitude of compound Fermat numbers, the Goldbach theorem states that $\gcd(F_n,F_m)=1$ if $n\neq m$. So $F_n=p_1^{n_1}\cdots p_i^{n_i}$ and $F_m=q_1^{m_1}\cdots q_j^{m_j}$ and $p_k\neq{q_j},\forall{j,k}$. Thus $p_1=F_n^{1/n_1}p_2^{-n_2/n_1}\cdots p_i^{-n_i/n_1}$ are infinite and there is always an infinitude of primes! Do you know another proof mentioned in the literature, please? Thank you.
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Let $n$ be a positive odd integer which is not a multiple of $5$.  Prove that there are infinitely many integers of the form $111\ldots111$ that are multiple of $n$.  I learned about Euler's theorem and $\phi$ function, but I don't know how to solve this problem.
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Starting from 2, the sequence of sums of all consecutive primes is:  $$\begin{array}{lcl}2 &amp;=&amp; 2\\     2+3 &amp;=&amp; 5 \\     2+3+5 &amp;=&amp; 10 \\     2+3+5+7 &amp;=&amp; 17 \\     2+3+5+7+11 &amp;=&amp; 28 \\     &amp;\vdots&amp; \end{array} $$ If the $n^\text{th}$ prime is $P_n$, then we can write $S_n=\sum_{i=1}^n P_n$.  I conjecture that the sequence $S_n$ contains an infinite number of primes.  I doubt I'm the first person ever to think this, but I cannot find reference to the idea, nor can I conceive a proof, nor a disproof.  Computationally, it can be verified that     $$S_{13,932}=998,658,581=P_{50,783,012},$$ and the sequence shows no sign of slowing down.  Can you prove the sequence $S_n$ contains an infinite number of primes?
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Well, we all know the twin prime conjecture. There are infinitely many primes $p$, such that $p+2$ is also prime. Well, I actually got asked in a discrete mathematics course, to prove that there are infinitely many primes $p$ such that $p + 2$ is NOT prime.
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If $n = 51! +1$, Then find no of primes among $n+1,n+2,\ldots, n+50$  Really speaking, I don't have any clue ...
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Are there any theorems that give us any information about the prime factorization of some integer $n+1$, if we already know the factorization of $n$?  Recalling Euclid's famous proof for the infinity of the set of prime numbers, I guess we know that if $n = p_1 p_2 p_3$, then $n+1$ cannot have $p_1$, $p_2$, or $p_3$ as factors. But is there any way we could use the information about $n$'s factorization to determine something more precise about the factorization of $n+1$?
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I have thought a lot but am failing to arrive at anything encouraging.   First try: If this is to be proved by contradiction, then I start with the assumption that let $n$ be a number which is a sum of two numbers, of which at least one is prime. This gives $n = p + c$, where $p$ is the prime number and $c$ is the composite number. Also, any composite number can be written as a product of primes. So I can say, $n = p + p_1^{e_1}.p_2^{e_2}...p_k^{e_k}$. From this, I get $n - p = p_1^{e_1}.p_2^{e_2}...p_k^{e_k}$, but I have no clue what to do next.  Second try: For an instant let me forget about contradiction. Since $n &gt; 11$, I can say that $n \geq 12$. This means that either $p \geq 6$ or $c \geq 6$. Again I'm not sure what to do next.  Finally, consider that the number 20 can be expressed in three different ways: $17+3$ (both prime), $16+4$ (both composite), and $18+2$ (one prime and one composite). This makes me wonder what we are trying to prove.  The textbook contains a hint, "Can all three of $n-4$, $n-6$, $n-8$ be prime?", but I'm sure what's so special about $4, 6, 8$ here.
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I'm trying to understand this proof from Stein's Elementary Number Theory, and I understand the pairing of inverses but not the other direction. I have two questions:  $1).$ When the proof says, $l$ a prime divisor of $p$, and so $l&lt;p$ and $l \mid (p-1)!$, is this because $l \mid p \Rightarrow l \mid p!=p*(p-1)!$ and so $l \mid (p-1)!$ (by Euclid)?  $2.)$ Why is it that $p \mid ((p-1)!+1)$?
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Let $g(x)\in \mathbb{Z}[x]$, a nonconstant polynomial. Show that the set of primes $p$ such that $p\mid g(n)$ for some $n\in \mathbb{Z}$ is infinite.  I don't know how to start. I have tried asuming that the set is finite but i haven't reached any contradiction.
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I have been given the following question in my homework:      Show that for every $n \in \mathbb{N}$ there are at least ${\log_2log_2 (n-1)}$ primes in $[1,n]$   I think it is related to euclidean and induction but I am having difficulties developing my solution.   Thanks for your help (:
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Intuitively it's true, but I just can't think of how to say it "properly".  Take for example, my answer to the following question:     Let $p$ denote an odd prime. It is conjectured that there are infinitely many twin primes $p$, $p+2$. Prove that the only prime triple $p$, $p+2$, $p+4$ is the triple $3,\ 5,\ 7$.   And my solution:     Given an odd integer $n$, between the three integers $n$, $n+2$ and $n+4$, one of them must be divisible by $3$... Three possible cases are $n=3k$, $n+2=3k$, and $n+4=3k$. The only such possible $k$ that makes $n$ prime is $k=1$. In this case, given an odd prime $p$, either $p=3$, $p+2=3$, or $p+4=3$. This would imply that $p=3$, $p=1$, or $p=-1$. The only of these three that is prime is $p=3$, therefore the only three evenly distributed primes are $3$, $5$, and $7$.   Is there a "better" way that I can assert that one of the integers is divisible by 3? This feels too weak.