All Questions
6 questions
4
votes
2
answers
709
views
On the product $\prod_{k=1}^{(p-1)/2}(x-e^{2\pi i k^2/p})$ with $x$ a root of unity
Let $p$ be an odd prime. Dirichlet's class number formula for quadratic fields essentially determines the value of the product $\prod_{k=1}^{(p-1)/2}(1-e^{2\pi ik^2/p})$. I think it is interesting to ...
4
votes
0
answers
238
views
On the values of $\prod_{k=1}^{(p-1)/2}(e^{2\pi i/12}-e^{2\pi i k^2/p})$ for primes $p>3$
In a recent preprint, I investigated
$$S_p(x):=\prod_{k=1}^{(p-1)/2}(x-e^{2\pi ik^2/p}),$$
where $p$ is an odd prime and $x$ is a root of unity.
Motivated by Question 337879 and Question 338325, ...
34
votes
1
answer
843
views
Can we write each positive rational number as $\frac1{p_1-1}+\ldots+\frac1{p_k-1}$ with $p_1,\ldots,p_k$ distinct primes?
It is well-known that any positive rational number can be written as the sum of finitely many distinct unit fractions. This is easy since
$$\frac1n=\frac1{n+1}+\frac1{n(n+1)}\quad\text{for all}\ n=1,2,...
17
votes
2
answers
3k
views
Does the equation $241+2^{2s+1}=m^2$ have a solution?
Let $p$ be a prime congruent to $1$ mod. 8.
If $p= 17$ one has : $p+ 8 = 5 ^2$.
If $p= 41$ one has : $p+ 8 = 7 ^2$.
If $p= 73$ one has : $p+ 8 = 9 ^2$.
If $p= 89$ one has : $p+ 32 = 11 ^2$.
If $...
10
votes
4
answers
1k
views
The smallest solution to $2^{2k}-1=\text{powerful}$
Integer is powerful if all the exponents in its factorization are at least $2$.
Every powerful integer can be written in the form $a^2 b^3$.
For odd $k$, define $F(k)=2^{2k}-1=(2^k-1)(2^k+1)$.
This ...
10
votes
0
answers
205
views
Does the diophantine equation $\,\prod_{k=1}^n(p_k^{x_k}-1)=y^2\,$ have always at least a solution for $\,n\gt2\,$?
P.G.Walsh proved in this paper that the diophantine equation $\,(2^{x_1}-1)(3^{x_2}-1)=y^2\,$ has no solution in positive integers $\,x_1$, $\,x_2\,$ and $\,y$.
If we generalize the previous equation ...