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1.)Prove that n^3-n is divisible by 6 using the exhaustion method.

2.) Prove that n^5 is divisible by 30 using the mathematical inductive method.

3.) Prove that the square of any integer is of the forms of 3k+1, bur not of the form 3k+2.

Hello Richard

(1) The exhaustion method consists of trying all the possibilies

n(mod 6) n^3 - n (mod 6)

0 0

1 1-1 = 0

2 8-2 = 0

3 27 - 3 = 0

4 64 - 4 = 0

5 125 - 5 = 0

Every possibility gives zero, so the theorem is proved.

(2) Your second question is in error. It should read : n^5 - n is div by 30.

Suppose the theorem is true for n = k.

Then k^5 -k is divisible by 30.

(a) It is true for k = 1 because 1^5 - 1 =0 is divisible by 30.

(b) Now we must prove (k+1)^5 - (k+1) is divisible by 5.

(k+1)^5 - (k+1) = k^5 + 5k^4 + 10k^3 + 10k^2 + 5k + 1 -k-1

We can now subtract k^5 -k , a multiple of 30.

(k+1)^5 -(k+1) = 5k^4 + 10k^3 + 10k^2 + 5k (mod 30)

= 5k(k^3 + 2k^2 + 2k +1) + (mod 30)

= 5k(k+1)(k^2 + k +1) (mod 30)

Now we examine the factors. 5 is a factor, 2 is a factor since we have the product of consecutive numbers. And 3 is a factor, since if k is not a multiple of 3,

either k+1 or k^2 + k + 1 is. (exhaustion method).

So that proves it is a multiple of 30.

So by mathematical induction, the theorem is true.

This was rather harder than the average inductive proof, so don't worry because you couldn't do it.

(3) I'm going to let you try that one for yourself and get back to me if you can't do it.

Hint: An integer is of the form 3k, 3k+1, or 3k+2. Square each of theses and reduce (mod 3).

You will be able to prove it by exhaustion!

Best wishes for your studies.

vijilant.

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