An interesting Math question

Can all the numbers be raised to different powers?

or is it 5 the power they are raised by?

There are no known solutions. I found this:

at:
http://mathworld.wolfram.com/DiophantineEquation5thPowers.html

Aren't you supposed to add like terms?

4^5 + 6^5 = 7^5 + 3^5

Because:

4^5 + 6^5= 10^5

and

7^5 + 3^5 = 10^5

so

10^5 = 10^5

there is actually a very simple solution to this problem.
We have: a^5+b^5 = c^5+d^5 (1)
Therefore: a^5+b^5 - (c^5 + d^5) = 0
We can see that this has the form of w + x - (y + z) = 0 (2)
Now let us take a simple solution that satisfies this equation e.g:
1 + 6 - (3 + 4) = 0 => 7 - 7 = 0

but a^5 = 1 therefore a = 1^(1/5) = 1
b^5 = 6 therefore b = 6^(1/5) = 1.431.....
c^5 = 3 therefore c = 3^(1/5) = 1.246......
d^5 = 4 therefore d = 4^(1/5) = 1.32......

These are positive solutions for a, b, c, d. Choose any values w,x,y,z for (2) that satisfy the equation and you can find an ifinite number of values for a,b,c,d in (1) as w,x,y,z are only bound by w+x = y+z.

This is not correct.
A^y + B^y does NOT equal (A+B)^y.
As a simple example, try 2^2 + 3^2.
2 squared is 4.
3 squared is 9.
4 + 9 = 13.
(2+3) squared = 5 squared = 25. (not 13).

AttemptATry:

Did you not see this:
and a,b,c,d are all unequal positive integers (so nonzero).