Special 4-Digit Square
A four digit integer a b c d satisfies the equation ( a b c d ) 2 = w x y z a b c d where a , b , c , d , w , x , y , z are not necessarily distinct digits and a = 0 , w = 0 . What is the value of w x y z ?
The answer is 8790.
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2 solutions
@Daniel Liu, sorry for this, but I just checked one of my programs for 4-digit automorphic numbers and found that 9376^2 = 89709376, and so I got the answer. :D
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Proof of d=5 does not work, I will give shortly. In fact I had typed it but by mistake delete it!!
We see that a b c d × a b c d = w x y z a b c d .
Subtracting a b c d from both sides, we get a b c d ( a b c d − 1 ) = w x y z 0 0 0 0 . Thus,
1 0 0 0 0 ∣ a b c d ( a b c d − 1 )
In addition, since a b c d and ( a b c d − 1 ) are consecutive integers, one must be even, and the other must be odd. Hence, either
[ 6 2 5 ∣ a b c d 1 6 ∣ ( a b c d − 1 ) or
[ 1 6 ∣ a b c d 6 2 5 ∣ ( a b c d − 1 )
We can substitute x = a b c d to make notation easier.
Case 1
In this case, 6 2 5 ∣ x and 1 6 ∣ x − 1 . We let x = 6 2 5 k . Thus 1 6 ∣ 6 2 5 k − 1 , or 6 2 5 k − 1 ≡ 0 ( m o d 1 6 ) .
This is simplified to k − 1 ≡ ( m o d 1 6 ) , where k = 1 is the obvious answer.
However, when we plug back in k = 1 to solve for x , we find that x = 6 2 5 , which is only a three-digit number. Thus, this case yields no solutions.
Case 2
In this second case, 1 6 ∣ x , so 6 2 5 ∣ x − 1 . We let x − 1 = 6 2 5 k . Thus, 1 6 ∣ 6 2 5 k + 1 , or 6 2 5 k + 1 ≡ 0 ( m o d 1 6 )
This can be simplified to k + 1 ≡ ( m o d 1 6 ) , at which k = 1 5 is an obvious solution.
Thus, x − 1 = 6 2 5 ∗ 1 5 = 9 3 7 5 , or x = a b c d = 9 3 7 6 .
Finally, ( 9 3 7 6 ) 2 = 8 7 9 0 9 3 7 6 (we confirm the last 4 digits of the result is the same as the squared number) and we get an answer of 8 7 9 0 .