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Prime numbers, diophantine equations, diophantine approximations, analytic or algebraic number theory, arithmetic geometry, Galois theory, transcendental number theory, continued fractions
0
votes
Why are the only numbers $m$ for which $n^{m+1}\equiv n \bmod m$ also the only numbers such ...
we can write:
$\displaystyle\int_{0}^{m}{x^m}dx\le\sum_{n=1}^{m}{n^m}
\le\int_{0}^{m+1}{x^m}dx$
so $\displaystyle\frac{m^{m+1}-(m+1)-m(m+1)k}{m+1}\le\sum_{n=1}^{m}{n^m}-1-mk-\le\frac{(m+1)^{m+1}-(m+1 …
2
votes
4
answers
686
views
solution of the equation $a^2+pb^2-2c^2-2kcd+(p+k^2)d^2=0$
i am wondering if there is a complete solution for the equation $a^2+pb^2-2c^2-2kcd+(p+k^2)d^2=0$ in which $a,b,c,d,k$ are integer(not all zero) and $p$ is odd prime.
0
votes
Boolean Cube of Primes
According to ingham's theorem $p_{n+1}-p_n< p_n+Ap_n^{5/8}$ in which $A$ is constant number.
Now let $p_n$ be largest prime less than or equal than $N$ so
$p_n\le N< p_{n+1}< p_n+Ap_n^{5/8}\le N+AN^ …