This time drk's solution was shorter than mine.  Essentially he
took square roots of both sides of



getting



Ssince the right side is a positive integer, the left side must 
be also.  That means there must be another factor of 11 under 
the square root radical, and the only possibility of the sum of 
two digits being a multiple of 11 is for (t+u) to be 11, and that 
is only true when t=6 and t=5. 
 
Here was my longer solution:
-----------------------------

There must exist integer k so that:



Factor the left side as the difference of two perfect squares:



Simplifying:











Write 99 as a product of primes:



Since the left side is an integer, the right side must be also.
So  must contain factors  as well as 
possibly another perfect square .

So  must be a perfect square of the form ,
where m is a positive integer. 



That gives:



The largest possible value of (t-u)(t+u) is 
when t=9 and u=0, or (9-0)(9+0)=81 and the smallest
is when t=1 and u=0, or (1-0)(1+0)=1.  Therefore the
right side must satisfy this:



 since 4 is the largest perfect 
square that does not exceed , and since
1 is the smallest largest perfect square that exceeds
.
Therefore


So m is either 1 or 2.

When m = 2, we have



The only ways 44 can be written as the product of two
integers are 1*44, 2*22, and 4*11. But the larger factor,
t+u cannot be more than 9+8 or 17, thus that would leave
only 4*11.  4 is even and 11 is odd. However t-u and t+u 
are either both even or both odd.  So we have ruled out 
m = 2.

That leaves only the possibility 

m = 1 and



Since 11 is prime, the only two positive integer factors it has
are 1 and 11.  Therefore the smaller factor  must be 1
and the larger factor 11.  So we have the system of equations:



Solving that gives us t=6 and u=5, which both can be digits.  
So the only solution is the two digit number 65.

Edwin