Find all Four digit number $\overline{abcd}$ such that $\overline{abcd} = a^x+b^x+c^x+d^x$ Find all the possible four digit numbers $\overline{abcd}$ such that $$\overline{abcd} = a^x+b^x+c^x+d^x$$
for a positive integral value of $x$ $(x>0)$.
My Take : 
Taking $x = 1$, We get $$\overline{abcd} = a+b+c+d$$  which is not possible
Similarly it is easy to show that there is no solutions for $x=2$ and $x=3$.
But I can't figure out how to solve for higher cases .Using a computer , I have been able to find some values of $\overline{abcd}$ but can we mathematically derive them?
 A: Not sure about the analytical description, but finding all of these numbers is not a very difficult task.
Easy to see that $x \leq log_2(10000)$ (if all digits are 0 or 1, then there is obviously no solution, and if there is at least one digit $s$ is 2 or more, then $s^x > 9999$). So it’s enough to loop through all possible options, and get all such numbers (1634, 4150, 4151, 8208, 9474).
A: If $d$ is odd, then an odd number of the powers need be odd. Otherwise, their sum won't be odd, this means we have an odd number of odd digits.
If $d$ is even, we need an even number ( including possibly 0) of the powers to be odd, forcing the digits they represent to be odd. 
These two observations, allow you to test only 4500 of 9000 numbers.
Looking at remainders on division by 3, we can show that any number divisible by 3, only takes on forms where an odd number of digits not divisible by 3, or all digits divisible by 3 if they could work at all for even powers. 
For odd powers, they need an equal mix of digits taking remainders 1 and 2 on division by 3, with an even number of digits divisible by 3 ( including 0). etc. 
