Sum Cool Squares.

First, let $x = \overline{ab}$ and let $y = \overline{cd}$ . $\overline{abcd}$ then becomes $100x+y$ , so the equation of interest is

$x^{2}+y^{2}=100x+y$

Now let's complete the square: $4x^{2}+4y^{2}=400x+4y$ $4x^{2}+4y^{2}-400x-4y=0$ $4x^{2}-400x+10000+4y^{2}-4y+1=10001$ $(2x-100)^{2}+(2y-1)^{2}=10001$

So the problem comes down finding a pair of numbers (a,b) such that the sum of the squares is 10001. One such pair is (100, 1), but that pair does not yield any values of x and y within the range 10-99. The other, less obvious pair is (76,65). Setting $2x-100=76$ yields x = 88, while setting $2y-1=65$ yields y = 33. So one of the desired 4-digit numbers is $8833$ .

To find the other, note that the fact that $76^{2}+65^{2}=10001$ holds true even if we take 76 and/or 65 to be negative. So let's solve for x and y again. $2x-100=-76$ yields x = 12, while $2y-1=-65$ yields y = -32. Clearly y = -32 doesn't work, so we can just use y = 33 again. This yields the other desired 4-digit number, $1233$ .

And of course, $8833+1233=\boxed{10066}$

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    s=0    
    for a in range(1,10):
         for b in range(1,10):
            for c in range(1,10):
                for d in range(1,10):
                   x=1000*a+100*b+10*c+d
                   if (((10*a+b)**2+(10*c+d)**2)==x):
                                  s=s+x
     print(s) 

The following python code gives the required solution = $\boxed{10066}$

{ for(i=1000;i<=9999;i++) {

            int k=i%100;
            sum=(k*k);
            j=i/100;
            a=(j*j);
        b=sum+a;
            if(b==i)
                 { 
                       System.out.println(i);
                 }
        }

the numbers are found to be 1233 and 8833 so sum=(1233+8833)=10066

The answer is 10066.

Substituting $\overline{ab}=x$ and $\overline{cd}=y$ we have $\overline{abcd}=100x+y$ . Writing our equation in terms of $x$ and $y$ , we have $x^2+y^2=100x+y$ making a quadratic in $x$ , we have $x^2-100x+(y^2-y)=0$ since $x$ and $y$ are positive integers, the discriminant must be a perfect square. So, $10000-4y^2+4y$ must be a perfect square. This implies that this expression must not end in $2,3,7,8$ . This implies that $y$ must not end in $1,7,4,9,2$ . Also, since $10000 > 4y^2-4y, y<50$ . Now this problem is a lot easier. We will move forward to make it simpler. Let $y$ end in $6$ . Then, the term $4y^2-4y$ will end in only one zero. But, a perfect square cannot end in an odd number of zeros. So, $y$ should not end in $6$ also. Similarly, for $5$ the only value where $4y^2-4y$ will end in two zeros is $y=25$ , but that will not give $10000-4y^2+4y$ to be a perfect square. So, $y$ must not end in $5$ also. Now, $y$ can end in only $3$ and $8$ . Also, $y$ starting digit can be at max $4$ . So, we have only 8 numbers to case-bash. And hence, we will finally get $y=33$ which gives $x=88(or)12$ . So, the sum of the two numbers is $1233+8833=\boxed{10066}$

I hope you like the method!!!

Python 2.7:

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def goal(n):
    number = str(n)
    return n == int(number[0:2])**2 + int(number[2:4])**2
total = 0
for n in xrange(1000,10000):
    if goal(n):
        total += n
print "Answer:", total