# Sums of powers of digits

Take a number written in decimal, like $25$. Take the sum of squares of its digits: $2^2 + 5^2 = 29$. Can you ever get the number you started with?

In fact, no positive natural number greater than one is the sum of squares of its decimal digits. However, 75 is pretty close: $7^2 + 5^2 = 74$. Numbers greater than 1000 can't be a sum of squares of their digits because $9^2 + 9^2 + 9^2 = 243$. By checking all possibilities, you can verify that no natural number greater than one is the sum of squares of its decimal digits.

The situation is different for cubes. First, take a look at this graph of the sum of cubes of digits minus the original number:
You can't actually tell here, but there is a number that is the sum of cubes of its decimal digits. It's $1^3 + 5^3 + 3^3 = 153$. For fourth powers there is one too: $9^4 + 4^4 + 7^4 + 4^4 = 9474$. The number 4150 works for fifth powers, and 548834 works for sixth powers. For $k > 2$, is there a number that is the sum of its $k$th powers? If not, are there infinitely many $k$?