Solution
= Solution
The strict inequality means $v_3(x^2+2)>2$. Taking $x=5$ gives $x^2+2=27$, so
$$
\boxed{|5^2+2|_3=3^{-3}<3^{-2}.}
$$
For the other expression, if $3\nmid x$ then $x^2+3\not\equiv0\pmod3$ 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 \b[no such <integer> exists] for the second strict inequality.