Intereting Posts

Closed form of $\int_0^1(\ln(1-x)\ln(1+x)\ln(x))^2\,dx$
Definite Integral $\int_0^{\pi/2} \frac{\log(\cos x)}{x^2+\log^2(\cos x)}dx = \frac{\pi}{2}\left(1-\frac{1}{\log 2}\right)$
If $|A|=30$ and $|B|=20$, find the number of surjective functions $f:A \to B$.
Does every automorphism of a permutation group preserve cycle structure?
What is Jacobian Matrix?
Let R be a relation on set A. Prove that $R^2 \subseteq R <=>$ R is transitive $<=> R^i \subseteq R ,\forall i \geq 1$
Is a mild solution the same thing as a weak solution?
Approximate spectral decomposition
Prove or find a counterexample: For all real numbers x and y it holds that x + y is irrational if, and only if, both x and y are irrational.
Why is a genus 1 curve smooth and is it still true for a non-zero genus one in general?
Number of solutions of $x_1+x_2+\dots+x_k=n$ with $x_i\le r$
Example of a group where $o(a)$ and $o(b)$ are finite but $o(ab)$ is infinite
Equation of line in form of determinant
Formal System and Formal Logical System
What is the exact value of $\eta(6i)$?

Show that every prime $p>3$ is either of the form $6n+1$ or of the form $6n+5$, where $n=0,1,2, \dots$

- Are 14 and 21 the only “interesting” numbers?
- Prime numbers of the form $(1\times11\times111\times1111\times…)-(1+11+111+1111+…)$
- A puzzle with powers and tetration mod n
- HCF/LCM problem
- Legendre's Proof (continued fractions) from Hardy's Book
- Prove the congruence $ \sum_{r=1}^{p-1}{(r|p) * r } \equiv 0 \pmod p.$
- Proving prime $p$ divides $\binom{p}{k}$ for $k\in\{1,\ldots,p-1\}$
- GCD proof - by contradiction?
- Prove that if $p$ is an odd prime that divides a number of the form $n^4 + 1$ then $p \equiv 1 \pmod{8}$
- $(m, m+2)$ is twin prime, iff $4((m-1)! + 1) \equiv -m \pmod {m(m+2)}$

Every integer is of the form $6n$ or $6n+1$ or $6n+2$ or $6n+3$ or $6n+4$ or $6n+5$ for some integer $n$. This is because when we divide an integer $m$ by $6$, we get a remainder of $0$, $1$, $2$, $3$, $4$, or $5$.

If an integer $m>2$ is of the form $6n$ or $6n+2$ or $6n+4$, then $m$ is even and greater than $2$, and therefore $m$ is not prime.

If an integer $m>3$ is of the form $6n+3$, then $m$ is divisible by $3$ and greater than $3$, and therefore $m$ is not prime.

We have shown that an integer $m>3$ of the form $6n$ or $6n+2$ or $6n+3$ or $6n+4$ cannot be prime. That leaves as the only candidates for primality greater than $3$ integers of the form $6n+1$ and $6n+5$.

**Comment**: In fact, it turns out that there are infinitely many primes of the form $6n+1$, and infinitely many primes of the form $6n+5$. Showing that there are infinitely many of the form $6n+5$ is quite easy, it is a small variant of the “Euclid” proof that there are infinitely many primes. Showing that there are infinitely many primes of the form $6n+1$ requires more machinery. But your question did not ask for such a proof.

$6$ divides $6n$, $2$ divides $6n+2$, $3$ divides $6n+3$, $2$ divides $6n+4$, and there are no other cases.

This is elementary algebra. For what value(s) of $n$ is $6n$ prime? $6n+2$? $6n+3$? $6n+4$? Are there any other possibilities besides these and the two that you already mentioned?

Indeed, all primes greater than $3$ are in the form of $6n-1$ and $6n+1$. I’ve studied this a few years ago. Here’s a basic visual proof of that using a sieve and isolation method that I used:

First, list down all the numbers in 6 columns:

$$\begin{array}{c|c|c|c|c|c}

1&2&3&4&5&6\\

7&8&9&10&11&12\\

13&14&15&16&17&18\\

\end{array}$$(the list will be infinite)

Then, we cross out the column of $2, 3, 4$, and $6$ as they are all composite. so we are just left out with two columns of $1$ and $5$. Using Algebraic progression with a difference of $6$, Column 1 generates the prime path of $$6n+1\quad[7,13,19,\ldots,\infty]$$ and column 5 generates the prime path of $$6n-1\quad[5,11,17,23,\ldots,\infty]$$

Thus, by isolation and sieve method, we can see that all primes must be in the form $6n-1$/$6n+1$.

Copied from Are all primes (past 2 and 3) of the forms 6n+1 and 6n-1?

[Considering] $n = 6q + r$

where q is a non-negative integer and the remainder $r$ is one of $0, 1, 2, 3, 4$, or $5$.

- If the remainder is $0, 2$ or $4$, then the number $n$ is divisible by $2$, and can not be prime.
- If the remainder is $3$, then the number $n$ is divisible by $3$, and can not be prime.
So if n is prime, then the remainder r is either

- $1$ (and $n = 6q + 1$ is one more than a multiple of six), or
- $5$ (and $n = 6q + 5 = 6(q+1) – 1$ is one less than a multiple of six).

- Discontinuity of Dirac Delta distribution
- Martingale theory to show f(x+s) = f(x)
- How do I prove that $\lim_{x→0} x⋅\ln x=0$
- Trouble understanding proof of this proposition on contact type hypersurfaces
- How do we prove that something is unprovable?
- Linear Dependence Lemma
- The other ways to calculate $\int_0^1\frac{\ln(1-x^2)}{x}dx$
- Exercise from Geometry of algebraic curves by Arbarello, Cornalba, Griffiths, Harris
- Proof – Inverse of linear function is linear
- If $\psi h$ is in $L^2$ for all $h\in L^2$, must $\psi$ be essentially bounded?
- Is it necessary for the homsets to be disjoint in a category?
- Improving bound on $\sqrt{2 \sqrt{3 \sqrt{4 \ldots}}}$
- Mathematical equivalent of Feynman's Lectures on Physics?
- Find the area of the region which is the union of three circles
- For what value of h the set is linearly dependent?