Double Or Nothing!
It's said that a number N with 4 digits is a DOUBLE SQUARE number when it equals the sum of the squares of two numbers: one formed by the first two digits of N, in the order they appear in N and the other formed by the two last digits of N in the order they appear in N.
For example, 1233 is a double-square number since 1233 = 12^2 + 33^2.
Find another 4- digit double-square number.
The answer is 8833.
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1 solution
Place the problem into an equation--
x^2 + y^2 = 100x + y, this is one equation for two variables. One solution to this equation is 1233. If you replace x = 12 you'll end up with--
y^2 - y - 1056 = 0, this case leads nowhere since delta < 0. But if you replace y = 33, then you'll have--
x^2 - 100x + 1056 = 0, this case gives you the problem's suggested solution 1233 and another one since delta > 0, which is 88 composing the answer 8833.
Another way to solve--- Let abcd be N
(10a+b)^2 + (10c+d)^2 = 1000a+100b+ 10c+d.
Of the LHS expression 100a^2 +20ab+b^2 + 100c^2 + 20 cd + d^2, only b^2 +d^2 can lead to the unit digit d cos 100a^2, 100c^2, 20ab, 20cd will have 0 as their last digit.
So (b,d) can be (2,3), (8,3), (8,8)
Substituting those values in equation and solving you will get 100(a^2+c^2) is approx equal to 1000a.
So (a,c) can be (2,4), (5,5), (3,5), (6,5), (7,5), (8,3), (8,4), (9,3)
Substituting for each, N = 8833