x^5-x can be divided with 30?

Question

How do you prove that?

Yes I am talking about x as a natural number
Any x

Answer 1

What you mean is that for any integer x, x^5 - x is divisible by 30, the subject being, not high-school algebra but number theory.
Every number is a multiple of 6 plus one of {0,1,2,3,4,5}.
[6x + {0,1,2,3,4,5}]^5 has what remainders when you divide by 6? For example 2^5 = 32 = 6*5 + 2. In other words
(6x + 2)^5 = 6z + 2, for some z. You will find that if you start with any number {0,1,2,3,4,5} you will have the same as remainders. So if you subtract them, X^5 - X will have remainders of zero. This means that X^5 - X is always divisible by six.
You can do the same thing with (5x + {0,1,2,3,4})^5 and you will find out the same thing; that all numbers of the form
x^5 - x are divisible by 5. Therefore it is also divisible by 30.

See Fermat's "little theorem" which applies here to the "divisible by five" part of the above argument.

Answer 2

x^5 - x = x(x^4 - 1). If x = 30 then the original statement is divisible by 30.

Answer 3

x = 2

x^5 - 2 = 32 - 2 = 30

Answer 4

Anything can be divided by 30.

Are you asking if you get a whole number or a rational number or what?

Answer 5

huh?

x^5-x=
x(x^4-1)

now if x was 2, then x(x^4-1) would equal 30, i'm not sure what you're looking for

Answer 6

Mathematical induction.

Answer 7

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