I recently stumbled upon a fact that all prime numbers past $3$ are of the form either $6n-1$ or $6n+1$.  Is it true? at least for numbers less than $10^9$.  And does it cover all primes?
82022
Show that every prime $p&gt;3$ is either of the form $6n+1$ or of the form $6n+5$, where $n=0,1,2, \dots$
121918
we all know that primes are special numbers, but let me ask you one thing that 'if we change the base of our widely used decimal to something else, lets say to octal, would the distribution of primes also change; or it is irrespective of the number system we use ?'  also if more basically we adopt binary number system which numbers would we call primes and how would sieve work ?  ask me if you are still not getting my question
72873
Find all prime numbers $p$ such that $p!+p$ is a perfect square.  I think the only ones are when $p!+p=p^2$, i.e. $p\in \{2,3\}$.  Any ideas at all?
52795
Let $n$ be an integer, prove that every prime factor of $12x^2+1$ has the form $6m+1$, where $m$ is an integer.  I've gone as far as knowing that all primes $p&gt;3$ are either $6m+1$ or $6m-1$, how do I use this, not sure whether to even use that.  I know about the Legendre symbol and quadratic reciprocity
23816
could any one tell me how to find the sum of all positive integers less than $n$ and relatively prime to $n$? $n&gt;1$  For $n=7$, I have  $\phi(n)=6$ and  the sum is ${6(6+1)\over 2}={n\over 2}.\phi(n)$, is that true for any $n$?
34210
Are all prime numbers greater than 2 odd?   I wasn't allowed to assume this was true.   AFSOC that there exists an even prime $n$. Then by definition of even, $n = 2q$, where $q$ is a positive integer. But if $n$ is even, it is divisible by $2$, hence that contradicts $n$'s prime-ness. QED.   What? Is there some wacky prime that can't be assumed odd?
66947
There should be infinitely many primes of the form $5+6n$. How do you prove it? The same should be true for $7+6n$.
70464
Is the below formula for the $n^{th}$ prime number in elementary functions useful somehow?  $$p(n)=\sum _{a=2}^{2^n} \sin \left(\pi  2^{\left(n-\sum _{b=2}^a \frac{\sin ^2\left(\frac{\pi }{b}((b-1)!)^2\right)}{\sin ^2\left(\frac{\pi }{b}\right)}\right)^2-1}\right)\frac{a \sin ^2\left(\frac{\pi }{a} ((a-1)!)^2\right) }{\sin ^2\left(\frac{\pi }{a}\right)}.$$
98853
Prove: Given a prime $p$, there are infinitely many $n\in \mathbb{Z}^+$, for   which $pn+1$ is a prime.   This is a simplified version of , so is there any elementary solution to this problem?  A proof by contradiction is recommended. Just like the proof of the existence of infinitely many primes by Euclid.
92419
Someone claimed that a number, multiplied by the number after it plus 17 is always prime, and showed several cases. I'm not a complete amateur in Number Theory, and I know that $17*18+17=17*19$, so it does not work for $n\equiv0(\mod17)$ but does it always work for other $n$ values? If not, can someone give me a counter example that I can show that person so they can correct their statement?
119709
How can one prove that numbers of the form $4x^2+1$ can only be divided by primes of the form $4y+1$ (e.g. there is no $x$ for which $7$ divides $4x^2+1$)?   On a quick lookup, the statement is given as the first comment in  and of course, manual testing concurs with this
85848
Prime numbers are numbers with no factors other than one and itself.  Factors of a number are always lower or equal to than a given number; so, the larger the number is, the larger the pool of "possible factors" that number might have.  So the larger the number, it seems like the less likely the number is to be a prime.  Surely there must be a number where, simply, every number above it has some other factors.  A "critical point" where every number larger than it simply will always have some factors other than one and itself.  Has there been any research as to finding this critical point, or has it been proven not to exist?  That for any $n$ there is always guaranteed to be a number higher than $n$ that has no factors other than one and itself?
33715
Question: Are there infinitely many primes of the form $4n+3$ and $4n+1$?  My attempt: Suppose the contrary that there exist finitely many primes of the form $4n+3$, say $k+1$ of them: $3,p_1,p_2,....,p_k$  Consider $N = 4p_1p_2p_3...p_k+3$, $N$ cannot be a prime of this form. So suppose that $N$=$q_1...q_r$, where $q_i∈P$  Claim: At least one of the $q_i$'s is of the form $4n+3$:  Proof for my claim: $N$ is odd $\Rightarrow q_1,...,q_r$ are odd $\Rightarrow q_i \equiv 1\ (\text{mod }4)$ or $q_i ≡ 3\ (\text{mod }4)$  If all $q_1,...q_r$ are of the form $4n+1$, then $(4n+1)(4m+1)=16nm+4n+4m+1 = 4(\cdots) +1$  Therefore, $N=q_1...q_r = 4m+1$. But $N=4p_1..p_k+3$, i.e. $N≡3\ (\text{mod }4)$, $N$ is congruent to $1\ \text{mod }4$ which is a contradiction.  Therefore, at least one of $q_i \equiv 3\ (\text{mod }4)$. Suppose $q_j\equiv 3\ (\text{mod }4)$  $\Rightarrow$ $q_j=p_i$ for some $1\leq i \leq k$ or $q_j =3$  If $q_j=p_i≠3$ then $q_j$ | $N = 4p_1...p_k + 3 \Rightarrow q_j=3$ Contradiction!  If $q_j=3$ ($\neq p_i$, $1\leq i \leq k$) then $q_j | N = 4 p_1...p_k + 3 \Rightarrow q_j=p_t$ for some $1 \leq i \leq k$ Contradiction!  In fact, there must be also infinitely many primes of the form $4n+1$ (according to my search), but the above method does not work for its proof. I could not understand why it does not work. Could you please show me?  Regards
17257
my question is about number theory,about prime numbers:  here is the question:  prove that there exists infinitely many prime numbers  of the form 4k+1 and also of the form 4k+3  and also of the form 6k+5
74289
Prove that if a number $n &gt; 1$ is not prime, then it has a prime factor  $\le \sqrt{n}$.  My answer is that this is not always true, because you can pick a non-prime number that is greater than $1$ and its factor is equal to its square root.
87572
I have a question about Sieve of Eratosthenes,   I refer at the "simply version" :   If I have an $\boldsymbol{n}$ and I want the prime numbers up to n,  I search and delete multiply up to  $ \leqslant \sqrt n $ .   For example:  $ n = 28;   \sqrt n = 5,29 $  After 5 I'm sure that I haven't delete multiply but I will find only prime numbers.   Now my question isn't how does it work, and why it is work..  But my question is why $ \sqrt n $ ? and for example why not $ \log n $ (is an example for trying to pass my question) How do you get to think that the $ \sqrt n $ has the power in this context?  How do you get to think that the $ \sqrt n $ assures me that there aren't more multiples to be erased?  Is there an explanation? graphic, numerical, of any kind? Thanks!
112475
I am trying to prove that ($\forall \ n\in\mathbb{N}$) there exists a prime number $q$ such that $n &lt; q \le 1 + n!$  I have made a graph with $n=0$ through $n=10$ and found solutions to all of them looking for a pattern and I see that $n!$ gets enormous fast and it becomes quite obvious that there is a prime number in between them.    I have considered trying to prove by contradiction that $q$ does not exist on that interval, but I don't know where to go from that statement.  Could anybody help me figure it out?  I have been staring at it for hours and I can't figure out where to go.  Thank you.