# Proving an equation with unorthodox components?

For example, how would you prove this function using the (k+1) substitution method?

Prove: n^5 – n is divisible by 10 for all n is greater than or equal to 2.

Thanks in advance.

Favorite Answer

If n is even, then n is divisible by 2 and (n^4 – 1) is a number divisible by 5. A number that is a multiple of 2 TIMES a number that is a multiple of 5 will be result in a number that is divisible not only by 2 and 5, but divisible by 10 as well. That is how come the number n(n^4 – 1) is divisible by 10….

Now…. if n is “ODD”… then n is NOT divible by 2…. HOWEVER, the (n^5 – 1) component of n(n^5 – 1) IS divisible by 10…. and that is how come when n is an odd number… the entire n^5 – n number is divisible by 10…. even though the n is ODD… the (n^5 -1) is a number divisible by 10…

Try it out…. you’ll see what I’m talking about… =)

Initial case: n=2 -> (2)^5-2 = 30, which is divisible by 10

Assumption: for n=k, (k)^5-k is divisible by 10

Test: for n=k+1, is (k+1)^5-(k+1) divisible by 10?

Things you need to know:

1) If numbers X and Y are divisible by 10–that is, X=10a and Y=10b–then X+Y is also divisible by 10. Think about it: X+Y = 10a + 10b = 10(a+b)

2) For any number to be divisible by 10, it must be divisible by 5 and some even number.

Strategy:

1) Completely expand your test case: (k+1)^5-(k+1).

2) If that expanded expression is divisible by 10, and you subtract out k^5-k (which we are assuming is divisible by 10), then the expression that remains should ALSO be divisible by 10. In other words, [(k+1)^5-(k+1)]-[(k)^5-k] should be divisible by 10.

3) Factor the resulting polynomial (it will be to the 4th power, and this factorization will require polynomial long division). For something to be divisible by 10 it has to be divisible by 5 and at least one even number. After you finish factoring, you’ll see that the polynomial satisfies this rule. Can you see why?

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