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 Three Squares (Posted on 2007-05-03)
I'm looking for a 4 digit square number where its first two digits are a square as are the last two.

What is its square root?

Incidentally, I'm not interested in the square of zero.

 See The Solution Submitted by brianjn Rating: 3.0000 (3 votes)

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 Analytic Solution Comment 9 of 9 |

Let the square root of the four digit number possess the form
(10*A + B).

Then, by the given conditions:

(10*A + B)^2 = 100*C^2 + D^2, where both C and D are non zero positive integers, with 1<=C,D<=9

or, 100*A^2 + 20AB + B^2 = 100*C^2 + D^2

Now, max(20AB + B^2) = 20*9*9 + 9^2 = 1701

Thus, 100*(A^2 - C^2) = 0,1,2, ...., 17

But, if 100*A^2 = 100*C^2 + i, for i = 0, 1, 2, ..., 17

If A> C, then (10A+B)^2 cannot be a perfect square unless :
(A, C)= (1, 0); (2, 0); (3, 0), (4, 0)
As C= 0, in each of the four cases, this would transform (10A+B)^2 to a number having two or less digits. This is a contradiction.

Thus, A^2 = C^2, so that: 20*A*B + B^2 = D^2
Since D<=9, we must have: 20*A*B + B^2 = D^2 <= 81< 100 ....(*)

If A>=5, then D^2 = 20AB + B^2>= 100B + B^2> 100, for B>=1.
This is a contradiction, so that A<=4.

If A<=2, then since A, B<10, we must have:
20*A*B + B^2 <= 40*B + B^2 < 40*10 + 10^2 = 500< 1000.

Accordingly, A = 3 or 4

Case (i):
If, A = 3, then  20*A*B + B^2 = 60*B + B^2 < 100, is satisfied only when B = 1, since B>=2, would force 20*A*B + B^2 > 100.

Thus (A, B) = (3, 1) is a possibility.

However, actual calculation reveals that 31^2 = 961, which contradicts  the provisions of the given problem.

Case (ii):

If A = 4, then  20*A*B + B^2 = 80*B + B^2 < 100, is satisfied only when B = 1, since B>=2, would force 20*A*B + B^2 > 100.

Thus, (A, B) = (4,1), which is borne out by acual calculations, as:

41^2 = 1681

Consequently, four digit number is 1681 and its square root is 41.

Edited on March 31, 2008, 12:50 pm
 Posted by K Sengupta on 2008-03-31 12:43:47

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