@PedroTamaroff @robjohn @anon @MikeMiller I want to check if the equation $3x^2+5y^2-7z^2=0$ has a solution in $\mathbb{Q}_p, p \neq 2,3,5,7$.
There are $\frac{p+1}{2}$ possible values of $3x^2$ and $\frac{p+1}{2}$ possible values of $7−5y^2$ for $x,y \mod p$. Then by the pigeonhole principle we have that for some $x,y$, we have $3x^2=7−5y^2 \mod p$, which then gives a solution.
Does this suffice, to show that there is a solution in $\mathbb{Q}_p, p \neq 2,3,5,7$, or do we have to apply also Hensel's Lemma?
There are $\frac{p+1}{2}$ possible values of $3x^2$ and $\frac{p+1}{2}$ possible values of $7−5y^2$ for $x,y \mod p$. Then by the pigeonhole principle we have that for some $x,y$, we have $3x^2=7−5y^2 \mod p$, which then gives a solution.
Does this suffice, to show that there is a solution in $\mathbb{Q}_p, p \neq 2,3,5,7$, or do we have to apply also Hensel's Lemma?
For the lower bound, note that the probability of selecting the first three primes is
$$P\left(\omega \in \{2,3,5\}\right) = \frac14 + \frac18 + \frac1{32}=\frac{13}{32}$$
and the desired probability is obviously higher than that.
To get an upper bound, consider the probability of choosing $2$...
Note the simple fact the integrand is positive and
$$\int_{0}^{1}\dfrac{1}{x\ln{x}}\cdot \underbrace{\frac{x}{\log(1-x)}}_{\large\text{near 0 it behaves like $-1$}}dx\longrightarrow \infty$$
Q.E.D.
Let $I = \int_0^1 x \tan(\pi x) \log(\sin(\pi x)) dx$. We have
$$I= \int_0^1 \dfrac{x \tan(\pi x)}2 \log(1-\cos^2(\pi x))dx$$
Hence,
$$2I = -\sum_{k=1}^{\infty}\dfrac1k\int_0^1 x \tan(\pi x) \cos^{2k}(\pi x)dx \,\,\,\,\,\,\,\, \spadesuit$$
\begin{align}
\int_0^1 x \tan(\pi x) \cos^{2k}(\pi x)dx &...
