Prove if $a\mid b$ and $b\mid a$, then $|a|=|b|$ , $a, b$ are integers. Form the assumption, we can say $b=ak$ ,$k$ integer, $a=bm$, $m$ integer.
 Intuitively, this conjecture makes sense. But I can't make further step.
 A: From what you wrote, $a=akm$, so $a(1-km)=0$. Also $b=bmk$, so $b(1-km)=0$. Thus either $a=b=0$ (and hence $|a|=|b|$), or $mk=1$. The units in $\mathbb Z$ are of course only $+1$ and $-1$.
A: Just continue your idea! $a=bm$, and $b=ak$ implies $a=akm$ and thus $1=km$. Which can only hold for $k=m=\pm1$.
Upd: of course, under $a,b\ne 0$, which comes from the setting.
A: $$a /b$$ then $$|a| \le |b|$$
$$b / a $$ then $$ |b| \le |a|$$
thus
$$|a| = |b|$$
A: As the OP Wes noted, $a \mid b$ implies $b = ka$ for some integer $k$; likewise $b \mid a$ yields $a = bm$.  Then $a = kma$, or $a(km - 1) = 0$.  Since $\Bbb Z$ is an integral domain, this forces $km = 1$.  (Note that $a = 0$ is prohibited by the assumption that $a \mid b$.)  $km = 1$ forces $k = m = \pm 1$, whence $a = \pm b$, whence $\vert a \vert = \vert b \vert$.  QED.
Note Added in Edit: In the light of Hagen von Eitzen's excellent answer, I suppose I should say that it is possible to allow $a = 0$ in the definition of $a \mid b$, since for integers this means $b = ka$ for some $k$; then we would clearly have $b = 0$ as well.  The definitions I am used to rule out the case $a = 0$.
Hope this helps.  Cheerio, 
and as always, 
Fiat Lux!!!
A: Just for fun, here is essentially Hagen's proof, written out more formally, but such that it is obvious that it really holds for both directions: it really proves if and only if.
In this answer all variables range over the integers.
\begin{align}
& a | b \;\land\; b | a \\
\equiv & \qquad \text{"definition of $\;|\;$, twice"} \\
& \langle \exists k :: k \times a = b \rangle \;\land\; \langle \exists m :: m \times b = a \rangle \\
\equiv & \qquad \text{"merge the independent quantifications} \\
& \qquad \phantom{\text{"}}\text{-- so that we can manipulate the two equalities together"} \\
& \langle \exists k,m :: k \times a = b \;\land\; m \times b = a \rangle \\
\equiv & \qquad \text{"substitute the left equality into the right"} \\
& \langle \exists k,m :: k \times a = b \;\land\; m \times k \times a = a \rangle \\
\equiv & \qquad \text{"arithmetic: divide right equality by $\;a\;$"} \\
& \langle \exists k,m :: k \times a = b \;\land\; (m \times k = 1 \;\lor\; a = 0) \rangle \\
\end{align}
This seems the appropriate time for a case distinction.  In case $\;a \not= 0\;$, we continue
\begin{align}
\equiv & \qquad \text{"use $\;a \not= 0\;$; simplify"} \\
& \langle \exists k,m :: k \times a = b \;\land\; m \times k = 1 \rangle \\
\equiv & \qquad \text{"arithmetic: the integer factors of 1 are both 1 or both $-1$"} \\
& \langle \exists k,m :: k \times a = b \;\land\; (k = 1 \lor k = -1) \;\land\; m = -k \rangle \\
\equiv & \qquad \text{"logic: one-point rule on $\;m\;$"} \\
& \langle \exists k :: k \times a = b \;\land\; (k = 1 \lor k = -1) \rangle \\
\equiv & \qquad \text{"logic: distribution; split $\;\exists\;$ on $\;\lor\;$; one-point rule on $\;k\;$, twice"} \\
& 1 \times a = b \;\lor\; -1 \times a = b \\
\equiv & \qquad \text{"arithmetic: simplify"} \\
& a = b \;\lor\; -a = b \\
\end{align}
In case $\;a = 0\;$, we instead continue
\begin{align}
\equiv & \qquad \text{"use $\;a = 0\;$; simplify"} \\
& \langle \exists k,m :: k \times 0 = b \rangle \\
\equiv & \qquad \text{"simplify"} \\
& 0 = b \\
\equiv & \qquad \text{"logic/arithmetic: use $\;a = 0\;$ twice -- suggested by our goal"} \\
& a = b \;\lor\; -a = b \\
\end{align}
So for both cases we've proven the required $\;a | b \;\land\; b | a \;\equiv\; a = b \;\lor\; -a = b\;$.
