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### Primes $p$ such that $5$ is primitive root $\bmod{p}$ , where $p$ is a $321$ prime

joriki's answer gives a great analysis. Calculations with PARI/GP let me find a counterexample (maybe the first one). The prime $p = 3\cdot 2^{18123}-1$ apparently found by Ballinger in April 1998 is ...
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### Find the least prime $p$, such that for some positive integers $a$, $b$, $c$, $d$, $(ab + a - b)(bc + b - c)(cd + c - d)(da + d - a) = 2023^2p.$

This answer is a work in progress. It shows that, without loss of generality, any integral solution must have $c$ and $d$ even and $a$ and $b$ odd, and either $a=1$ or $b=1$ or both. It is quite a lot ...
$$\sum_{p\leqslant N}p^{-1-i}=\sum_{n=2}^N\big(\pi(n)-\pi(n-1)\big)n^{-1-i} \\=\pi(N)N^{-1-i}+\sum_{n=2}^{N-1}\pi(n)\big(n^{-1-i}-(n+1)^{-1-i}\big)$$ and, since $n^{-1-i}-(n+1)^{-1-i}=(1+i)n^{-2-i}+... 1 vote ### Find the least prime$p$, such that for some positive integers$a$,$b$,$c$,$d$,$(ab + a - b)(bc + b - c)(cd + c - d)(da + d - a) = 2023^2p.$If$a=b=c=1$then the equation reduces to$2d-1=2023^2p$and there are solutions for all odd$p$. So we can concentrate on$p=2$. $$(ab+a-b)(bc+b-c)(cd+c-d)(da+d-a)=2.7^2.17^4$$ For the LHS to be ... 1 vote ### Find the least prime$p$, such that for some positive integers$a$,$b$,$c$,$d$,$(ab + a - b)(bc + b - c)(cd + c - d)(da + d - a) = 2023^2p.$(Hm, this doesnt' work out as easily as I thought it would, so ignore for now.)$ab+a-b) = (a-1)(b+1) + 1$. So if$ab+a-b = 1$, this just requires$a = 1$(with no restrictions on$b$). Show that$p$... 5 votes Accepted ### Find all two-digit prime number pairs$p$and$q$, for which$pq+1$is a perfect square. We can assign$pq+1$to equal to another variable, say$k^2$. in other words, let $$pq+1=k^2$$ where$k$is a positive integer. Rearranging this equation yields $$pq=k^2 - 1$$ $$\implies pq = (k-1)(k+... 8 votes ### Find all two-digit prime number pairs p and q, for which pq+1 is a perfect square. Any pair of twin primes will work, because (n+1)(n-1)=n^2-1. Those are the only possible solutions because the problem is equivalent to saying n^2-1=(n+1)(n-1) is a product of two primes, so ... 1 vote Accepted ### Recalling a theorem from vague memory: A monoid, in some sense, cannot "describe" the language (over one letter?) of words of prime length. I guess you refer to the syntactic monoid of a language. See also the French entry of Wikipedia, which gives more details. A key property is that a language is regular if and only if its syntactic ... 1 vote Accepted ### Odd numbers form pairs infinitely often in this sequence? Consider the list of numbers which are double a prime.$$4,6,10,14,22,26,34,38,46, ...$$Then I(k) is odd if and only if k is an odd number between a 2i-1th and 2ith element in the above list. ... 2 votes ### Question on Brun's theorem on twin primes Warning: This is only a partial answer due to using a yet unproven (nor refuted) conjecture. Quite a bit too long to leave as a comment anyway. If the first Hardy-Littlewood conjecture for prime k-... 3 votes Accepted ### Question on Brun's theorem on twin primes Let N be some even integer. Then it is known that there exists some constant A>0 such that$$ \#\{p\le x:p,p+N\text{ prime}\}\le{Ax\over\log^2x}\prod_{\substack{p|N\\p>2}}{p-1\over p-2}. $$... 2 votes ### Is 12 the only number that has the same digits as its prime factorization? This is not a complete answer, but too long to fit in a comment. This shows that if k\geq100, then k must be divisible by the cube of a prime. There is plenty of room for improvement of the ... 1 vote ### How to prove that N!+p(N) can't be a perfect power, where p(N) is the N^{th} prime? For y even, we can easily prove a stronger statement:$$p(N)\text{ is a quadratic non-residue modulo }N!$$Proof: For N <= 9, check computationally and confirm the statement. Hence we can ... 0 votes ### Lucas-Lehmer test for Mersenne and Wagstaff numbers? If you look at the sequence in OEIS A029858 changing the index is equal to the sequence you used then you get$$S_{p-2}=\frac{3^{2^p-1}-3}{2}$$for Mersenne numbers but also$$S_{p-2}=\frac{3^{2^p-1}-... 0 votes ### Why are there always two primes between$2^n$and$2^{n+1}$? We have certain properties, that have been proven, and other properties, that are totally sure, but not proven. It has been proven that between$n$and$2 \times n$, we have at least 1 prime. It has ... 0 votes ### Why are there always two primes between$2^n$and$2^{n+1}$? By the prime number theorem, the prime counting function$\pi(n)$is quite close to $$\pi(n) \approx \frac{n}{\log n}$$ so $$\pi(2^{n+1}) - \pi(2^n) \\= \frac{2^{n+1}}{\log 2^{n+1}} - \frac{2^{n}}{\... 4 votes Accepted ### Why are there always two primes between 2^n and 2^{n+1}? This follows from a strengthening of Bertrand's postulate. In fact it can be shown that for every \varepsilon>0, for all sufficiently large n there exists a prime between n and (1+\... 0 votes ### Finding number with prime number of divisors? For prime factorization n = p_1^{e_1} \cdots p_r^{e_r} , the number of divisors is \sigma(n) = (e_1 + 1) (e_2 + 1)\cdots (e_r + 1). For that to be prime, n must be a prime power p^e where e + ... 1 vote ### Is F_n+F_{n^2+1} composite for all odd positive integers n>1? This is not an answer to your question, just a comment that would be too long. I have sieved n < 10000 (odd and even) up to p = 10^{11} and the candidates ... 5 votes Accepted ### Does for every integer k exist a positive integer n such that F_n+k is prime? No, f(k) is not total. Reference https://oeis.org/A361902 for k\ge 0, where it states that if k\equiv 14475 \pmod {1647030} then there is no solution. The OEIS page links to a proof by Robert ... 1 vote ### Find all relative prime positive integers p and q such that p+q=(p-q)^3. Let p and q be two positive integers satisfying$$p+q=(p-q)^3.$$Let r:=p-q so that q=p-r and so \begin{eqnarray} p&=&\frac{r^3+r}{2}=\frac12r(r^2+1),\\ q&=&\frac{r^3-r}{2}=\... 2 votes ### Find all relative prime positive integers p and q such that p+q=(p-q)^3. Let n=p-q. Then$$p=\frac{n^3+n}2,\quad q=\frac{n^3-n}2.$$If n is odd then it divides p,q (because n^2\pm1 is even). This is impossible because q would be equal to \frac{1^3-1}2=0. If n=... 2 votes Accepted ### Find all relative prime positive integers p and q such that p+q=(p-q)^3. Since p-q\mid p+q, we have p-q\mid (p+q)+(p-q) = 2p and p-q\mid (p+q)-(p-q) = 2q, hence p-q\mid 2\gcd(p, q) = 2. We divide cases. (i) p-q = 1: then p = q+1, so the equation reduces to 2q+... 1 vote Accepted ### Finding an x that satisfies the Jacobsthal function for the primorial of the nth prime. A list, up to 24, is given at https://oeis.org/A049300/b049300.txt Here it is:$$ \matrix {1& 2\hfill\cr 2 &2\hfill\cr 3& 2\hfill\cr 4& 2\hfill\cr 5 &114\hfill\cr 6 &9440\... 0 votes ### If$30$divides$p_1^4 + p_2^4 + \ldots + p_{31}^4$. Prove that$p_1=2$,$p_2=3$and$p_3=5$. Let$a=1$if$2$is among the$p_i$'s, and$a=0$else. Define similarly$b$for$3$and$c$for$5.$For any prime number$p$coprime to$30,$we have$p^4 ≡ 1\bmod2,3,5$hence$\bmod{30}.\$ Therefore, ...
If you want to analyze the numerator and denominator of the series $$H_n=1+\cdots +\frac{1}{n},$$ you can do so by noticing that $$H_{n+1}-H_{n}=\frac{1}{n+1}$$ and put $$H_n=\frac{N_n}{n!}$$ yielding ...