Past exam of the analytic number theory and elliptic curves course of the University of Oxford / 2007 / B / 5

The strict inequality means $v_3(x^2+2)>2$. Taking $x=5$ gives $x^2+2=27$, soFor the other expression, if $3\nmid x$ then $x^2+3\not\equiv 0(\mathrm{mod}3)$ and its P-adic valuation is zero. If $3\mid x$, then $x^2+3=3(3(x/3{)}^2+1)$ has P-adic valuation exactly one. It can never have valuation greater than two. Thus no such integer exists for the second strict inequality.

In the P-adic numbers, ${\sum }_{j\ge 0}{p}^{2j}$ converges to $1/(1-p^2)$. ConsequentlySince $p\ne 2$, the digits one and two are both between zero and $p-1$, so no carrying is required. These are its p-adic digits, with the unique negative-power digit at exponent minus one.

No prime works. The P-adic valuation induces a well-defined homomorphismThe class of $p$ maps to one, whereas the class of $p^p$ maps to zero. They therefore cannot be equal. Equivalently, $p$ cannot be a $p$th power because its valuation is not divisible by $p$.

Choose a nonzero integer $D$ with $D\alpha ,D\beta \in \mathbb{Z}$. By the Chinese remainder theorem, choose $z_n\in \mathbb{Z}$ satisfying $z_n\equiv D\alpha (\mathrm{mod}{q}^n)$ and $z_n\equiv D\beta (\mathrm{mod}{r}^n)$. Put $x_n=z_n/D$. ThenThis proves the requested simultaneous approximation, even when the rational targets have denominators divisible by either prime.

A rational sequence converging to zero at every standard absolute value is:Its ordinary absolute value is at most $1/n!$, hence tends to zero. For a fixed prime $p$ and all sufficiently large $n$, the denominator is congruent to one modulo $p$, so $v_p(y_n)=v_p(n!)\to \infty$. Therefore $|y_n{|}_p\to 0$ for each prime as well. The assertion concerns each fixed prime separately, rather than uniform convergence over all primes.