Meaning of congruence notation for Bernoulli Numbers I am studying Theorem 4(von Staudt's Theorem) in Borevich-Shafarevich's Number Theory(1966)(page 384) which states:

Let $p$ be a prime and $m$ an even integer. If $(p-1)\nmid m$, then
  $B_m$ is $p$-integral(that is, $p$ does not appear in the denominator
  of $B_m$). If $(p-1) | m$, then $pB_m$ is $p$-integral, and $$pB_m
\equiv -1 (mod\hspace{0.05in}p).$$

What's the meaning of this congruence? 
For instance for $p=3$ and $m=4,$ $3B_4=3(\frac{-1}{30})=\frac{-1}{10}.$ How is $$\frac{-1}{10} \equiv -1 (mod\hspace{0.05in}3) ?$$
Thank you!

The fraction $\frac{1}{n}$ is, by definition, the number which you can multiply $n$ by to get the answer $1$. Now the same definition works in congruence arithmetic. For example modulo $p$, as long as $n$ is not divisible by $p$ there is some number $m$ such that $nm \equiv 1$ mod $p$. In that case $\frac{1}{n}\equiv m$ (mod $p$).
There's a sign error in the statement of the theorem in that book. The formula for when $m|p-1$ (as you've now corrected) should be $pBm \equiv -1$ modulo $p$. 
So for $m=4$ and $p=3$ we get $pB_4 = \frac{-1}{10} \equiv -1$ modulo 3, which makes sense since $10\equiv 1$ (mod $3$).
For another example let's look at $m=16$ and $p=5$, and $B_{16} = \frac{-3617}{510}$. We have
$$pB_m = 5 \frac{-3617}{510} = \frac{-3617}{102},$$
Now $102 \equiv 2$ (mod $5$), and $3*2 = 6 \equiv 1$, so $\frac{1}{102} \equiv 3$. We also have $-3617 \equiv -7 \equiv 3$ (mod $5$) and so indeed
$$pB_m = \frac{-3617}{102} \equiv \frac{3}{102} \equiv 3 * 3 = 9 \equiv -1 \text{(mod }5)$$ 
