Positive real numbers $a$ and $b$ have the property that
\[\sqrt{\log{a}} + \sqrt{\log{b}} + \log \sqrt{a} + \log \sqrt{b} = 100\]
and all four terms on the left are positive integers, where log denotes the base 10 logarithm. What is $ab$?
$\textbf{(A) }   10^{52}   \qquad        \textbf{(B) }   10^{100}   \qquad    \textbf{(C) }   10^{144}   \qquad   \textbf{(D) }  10^{164} \qquad  \textbf{(E) }   10^{200}$
Since $\sqrt{\log{a}}$ is a positive integer, we get $\log a = x^2$ for some integer $x$; since $\log \sqrt{a} = \tfrac 12 \log a$ is a positive integer, we get $x=2m$. Thus $a=10^{4m^2}$; similarly $b=10^{4n^2}$. Substituting, we get $2(m+n+m^2+n^2)=100$, i.e. $m(m+1) + n(n+1) = 50$. It follows that $m,n \le 6$. The values of $m(m+1)$ for $m=1,\ldots , 6$ are 



$\qquad m$
$1\qquad$
$2\qquad$
$3\qquad$
$4\qquad$
$5\qquad$
$6\qquad$


$m(m+1)\qquad$

$2$

$6$

$12$

$20\qquad$

$30$

$42$

Two of those values must add up to $50$ and we see that $20+30=50$, so $m=4, n=5$ and  $ab=10^{4(m^2+n^2)}=10^{4(4^2+5^2)}$, and our answer is $\boxed{\textbf{(D) } 10^{164}}$.