Find the four digit number with perfect square Find the four digit number 


*

*whose first two digit is a perfect square

*and second two digit is a perfect square

*and the 4 digit number itself is a perfect square


for ex: 
if the number is "abcd" then
"ab" is a perfect square
"cd" is a perfect square
"abcd" is also a perfect square
what is the number? how to find it easily?
Edited
i need a algebra solution.
 A: The number is 1681.
It could only be one of a few numbers starting either 16cd, 25cd, 36cd etc. Then I just squared a few numbers until i could see the cd was a square number.
Further, the number cannot be > 2500, since 50^2 = 2500, and 51^2 = 2601, so there would not be a cd other than '00' following 25cd, 36cd etc
A: Let the number is $\overline{abcd}$. So you have $\overline{ab} = k^2$, $\overline{cd} = l^2$ and $\overline{abcd} = m^2$, where $0\leq k,l \leq 9$.
One then has $$m^2 = 100k^2 + l^2.$$
Then $$2^25^2k^2 = (m-l)(m+l).$$
Note that $m-l$ and $m+l$ are even. 
Case 1: $m+l = m-l = 10k$. Then $l=0$, and the number which you want to find are $1600$, $2500$, $3600$, $\dots$, $8100$.
Case 2: $m+l = 50$ and $m-l=2k^2$, then $2l = 50-2k^2$, or $l=25-k^2$, note that $0\leq l \leq 9$, so $k=4$ and $l=9$. So, the number is $1681$.
A: Why not just brute-force it? The problem is a pretty opaque and unnatural to begin with.
>>> min([i for i in range(1000,10000) if (i**0.5).is_integer() and (i/100)**0.5).is_integer() and ((i%100)**0.5).is_integer()])
1600
Unmin-ed version:
>>> [i for i in range(1000,10000) if (i**0.5).is_integer() and ((i/100)**0.5).is_integer() and ((i%100)**0.5).is_integer()]
[1600, 1681, 2500, 3600, 4900, 6400, 8100]
