[Micky76]

Basically I was attempting to do some examinations that are sat in Poland at the age of 18 (equivalent to A-level) and I encountered this question which is written very poorly. I have translated the question and one mark scheme answer method to the best of my ability. Here it goes:

Prove that for every integer k, k^6 - 2k^4 + k^2 is divisible by 36.

(Seems like they forgot to say k>1)

Answer:

k^6 - 2k^4 + k^2 = k^2(k^4 - 2k^2 + 1) = k^2(k^2 - 1)^2
=[k(k+1)(k-1)]^2

>Showing that k(k+1)(k-1) is divisible by 6:

(k-1)k(k+1)

For any three consecutive integers at least one is even and exactly one is divisible by 3 (Not true if k=1). The square of the product of these integers is divisible by 36.

Therefore, an integer given by k^6 - 2k^4 + k^2 where k is an integer is divisible by 36.

-End of answer-

Just wanted everyone thoughts on this, the question is basically disproved with -2<k<2 but do you think that it is an adequate "proof" for k outside of this range?

EDIT: I was wrong and the question and solution is right. Perhaps I was to critical about this question, sorry.

[notnek]

I do think it's an adequate proof. Are there any specific parts that you weren't sure about?

[Micky76]

(Original post by notnek)
I do think it's an adequate proof. Are there any specific parts that you weren't sure about?

Just that it is wrong for -2<k<2. But I assume that they just forgot to put that in.

[DFranklin]

You seem to think 0 is not divisible by 36. It is.

[The Polymath]

(Original post by Micky76)
Just that it is wrong for -2<k<2. But I assume that they just forgot to put that in.

It seems exactly like the proof by induction question we get in A-level Further Maths, have you tried doing it that way?

[Venomilys]

surely you can do this quickly

k = 36n

k^6 - 2k^4 + k^2 = k (k^5 - 2k^3 + k) = 36n (k^5 - 2k^3 + k) = 36p.

correct me if I'm wrong...

[notnek]

(Original post by Ilyas)
surely you can do this quickly

k = 36n

k^6 - 2k^4 + k^2 = k (k^5 - 2k^3 + k) = 36n (k^5 - 2k^3 + k) = 36p.

correct me if I'm wrong...

What you have shown is: "If k is divisible by 36 then k^6 - 2k^4 + k^2 is divisible by 36".

The statement in the OP is "For every integer k, k^6 - 2k^4 + k^2 is divisible by 36".

[Venomilys]

(Original post by notnek)
What you have shown is: "If k is divisible by 36 then k^6 - 2k^4 + k^2 is divisible by 36".

The statement in the OP is "For every integer k, k^6 - 2k^4 + k^2 is divisible by 36".

oops, sorry! should have read it properly.

[Micky76]

(Original post by DFranklin)
You seem to think 0 is not divisible by 36. It is.

You hit the nail on the head. That clarifies everything. Thank you.

[Micky76]

(Original post by DFranklin)
You seem to think 0 is not divisible by 36. It is.

Just one thing though, so it is technically divisible by any number except for zero itself.

[DFranklin]

Yes.