Oops, there was a typographic error in the formula at the start of the month. It is right now. Thank you to Andrei Lazanu of school 205 Bucharest for what follows.

First, I tried to solve the problem in the "conventional mathematical way", and I listed the powers of 1, 2, 3, 4, 9 from 0 to 9, looking in each case to eliminate those greater than 5 digit numbers, attributed to both

and

Because it seemed very long, I tried the problem from a "calculatoristic" point of view:

First I observed that N must be greater than 0, and each other number must have the values from 0 to 9. I wrote a program in MATLAB to calculate the numbers.

The idea of the program is to test for all digits the values from 0 (1) to 9. For this I wrote a cycle allowing $n$ to have values between 1 and 9. For each value of $n$ , I give, in an inside cycle, to $r$ values from 0 to 9. So, this works as follows: $n$ has the value 1, and for this $r$ has in turn values from 0 to 9. This is the idea of a cycle inside another one. Each cycle ends with end.

For this problem there are 5 cycles, the most interior being for $h$ , that has values between 0 and 9.

When I arrived at this cycle, I test if the condition of the problem:

$r^n-i+c^h=1000\times r+100\times i+10\times c+h$

is fulfilled. If yes, the number

$1000\times r+100\times i+10\times c+h$

is written on the screen.

Because this way I'll find all solutions, even with repeated digits, I must in a further step eliminate them.

Here is the program:

$\mathrm{for}n=1:9;$

$\mathrm{for}r=0:9;$

$\mathrm{for}i=0:9;$

$\mathrm{for}c=0:9;$

$\mathrm{for}h=0:9;$

$\mathrm{if}{r}^n-i+c^h-10000\times n-1000\times r-100\times i-10\times c-h==0$

$a=[nrich]$

$\mathrm{end}$

$\mathrm{end}$

$\mathrm{end}$

$\mathrm{end}$

$\mathrm{end}$

$\mathrm{end}$

The results given were the following:

a = 5 9 0 5 0

and

a = 5 9 2 6 3

As in the first one the digits are not distinct, this is not a solution.

The second is the unique solution of the problem:

n = 5
r = 9
i = 2
c = 6
h = 3