A *simple* maths problem.

[jack]

Q. Show that *=N⁴+N²+2N is a multiple of 4.
A. It's a multiple of N so changing N by four doesn't change the expression being a multiple of four or not[1]. So we only need to check N=0,1,2 and 3. But it works for those cases or you wouldn't have asked the question. QED.

Full credit? You would, I think, normally accept four one-line substitutions as sufficient? You don't actually write out the multiplications, you just say "1⁴+1²+2.1=4", and you could do the same in modular arithmetic and all the answers would be zero. And this is just a completely trivial collapse of that.

It's even realistic -- if you're proving S(n) for all n and find that S(n)∀n<N=>S(n)∀n [2] someone's probably crunched through N on a supercomputer already[3], and it each m must be S as if m wasn't they'd have published, the pleasure of a paper titled "Hahahahah!" and reading in its entirity "~S(m)" :)

But something about the answer bothers me. I think it's that (a) it's a smartarse and (b) a good habit to be into is to at least note down each case and put a tick by it to show you've tried them all in your head and didn't forget one.

[1] Yes, I could have used modular arithmetic language, but the colloqiual language fits the point better.
[2] Googling for "html math symbols" first produces a page suggesting you use the symbol font :(
[3] OK, professional problems are complicated enough that m might be too stupidly big. Witness the colouring theorem. But it's a major major major major[4] step, and someone else will hopefully fill in the blanks, even if they turn out to be hard.
[4] C22 reference.

Comments

[miss-next.livejournal.com]

No, there ought to be a more logical way to do it than that.

If N is even, then N^4, N^2 and 2N are automatically divisible by 4 (you can see that this is so by writing N = 2M, where M is an integer). Hence the result just pops out.

If N is odd, then you can write it as 2M + 1. Expanding the original expression gives you

(16M^4 + 16M^3 + 12M^2 + 4M + 1) + (4M^2 + 2M + 1) + 4M

which reduces to

16M^4 + 16M^3 + 16M^2 + 10M + 2

...

Ah. You need to prove that 10M + 2 is divisible by 4, but this follows only if M is odd, and I've almost run out of lunch hour.

Still, with a bit of luck this is a start!

[numberland.livejournal.com]

If N is odd, then you can write it as 2M + 1. Expanding the original expression gives you

(16M^4 + 16M^3 + 12M^2 + 4M + 1) + (4M^2 + 2M + 1) + 4M

Um that should be:
(16M^4 + 32M^3 + 24M^2 + 8M + 1) + (4M^2 +4M +1) + 4M + 2
Then it drops out

[miss-next.livejournal.com]

You're right - I was trying to do the algebra very quickly because I had limited time available. :-)

[cartesiandaemon.livejournal.com]

It happens. But see, my example that's not even a risk. Even if you write out the four special cases, it's still trivial and obvious and short imho :)

[ewx.livejournal.com]

Or observe that f(0)=0 and f(N+1)-f(N) = 4+6N+6N²+4N³ and 6N+6N² = 3.2.N(N+1) and one of N and N+1 must be even. Induct.

[cartesiandaemon.livejournal.com]

Induct twice to get negative numbers too (the difference is still the same), but yes. The multiplicity of easy ways of doing it was known and doesn't really affect the question I thought most amusing.

[ewx.livejournal.com]

Err, yes, I should have considered N<0. Hmm. Taking into account the fact that someone is asking a question would obviously be wrong in an exam, and in real life (or I suppose in an exam) they might be trying to trick you, like the king in the story asking his wise men why some false thing happens.

[cartesiandaemon.livejournal.com]

Exactly. You're generally supposed to be doing it as if you didn't know the answer :)

Is that a specific story about wise men, or just a general theme?

[ewx.livejournal.com]

I think I first heard it as a specific story but I think there's more than one flavour of it and I can't remember the details.

[cartesiandaemon.livejournal.com]

OK, thanks.