Math: Wufnu's Theorem

[Monte314]

I have concocted a theorem to honor the recently-arrived math guy, wufnu. It is a theorem about divisibility of integers having a certain form. In a few days, I will post the general form of Wufnu's Theorem, with a proof (if none has been provided by then):


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Hah! Made you look!

[enWTFp]

M(M^6 - 14M^4 + 49M^2 - 36) = M(M^2-1)(M^4 - 13M^2 + 36) = M(M^2-1)(M^2 - 4)(M^2 - 9) =
(M-3)(M-2)(M-1)M(M+1)(M+2)(M+3) which is divisible by 7! = 5040.

In general n(n+1)...(n+k-1) is divisible by k! , because it runs all the possible remainders over 1,2,..,k.
I enjoy a problem the most when it can be solved without any writing on paper, by just looking long enough at it. Thank you, Monte.

[Monte314]

M(M^6 - 14M^4 + 49M^2 - 36) = M(M^2-1)(M^4 - 13M^2 + 36) = M(M^2-1)(M^2 - 4)(M^2 - 9) =
(M-3)(M-2)(M-1)M(M+1)(M+2)(M+3) which is divisible by 7! = 5040.

In general n(n+1)...(n+k-1) is divisible by k! , because it runs all the possible remainders over 1,2,..,k.
I enjoy a problem the most when it can be solved without any writing on paper, by just looking long enough at it. Thank you, Monte.


Double Yay!

Not only has enWTFp solved the special case, but has also correctly stated the generalization and given an indication of its proof.

Anybody else?

[Wufnu]

Man, I never would have gotten that one. I probably should have paid more attention in abstract algebra D: That was the best part of that class. Some of the things students would do to prove something were downright brilliant and obscure as hell :P


"divisibility of integers" would have lead me down the right path, though.

[LordMaiestas]

I nvr even touch on this type of maths...

[Monte314]

Wufnus’s Theorem


Let M, N be non-negative integers with M>N.

Then the product (M*M-1*1) (M*M-2*2) (M*M-3*3)...(M*M-N*N) is divisible by (2N+1)!

(Note: recall that N! = N(N-1)(N-2)…(3)(2)(1), so, for example 5!=5x4x3x2x1 = 120)

Examples of Wufnu's Theorem:

For N=0, Wufnu says that M is divisible by 1 for every non-negative integer M.
For N=1, Wufnu says that M^3-M is divisible by 6 for every non-negative integer M.
For N=2, Wufnu says that M^5-5M^3+4M is divisible by 120 for every non-negative integer M.
For N=3, Wufnu says that M^7-14M^5+49M^3-36M is divisible by 5,040 for every non-negative integer M.

Etc.

Proof: (there are proofs that use only number theoretic notions, but they are more complicated)

The expression a*a – b*b can be factored (a + b)(a – b). Therefore, the product in the statement of the theorem can be written:

(M*M-1*1)(M*M-2*2)(M*M-3*3)...(M*M-N*N = M(M+1)(M-1)(M+2)(M-2)…(M+N)(M-N)

= (M-N) (M-(N-1)) (M-(N-2))…(M-2) (M-1) M (M+1) (M+2)…(M+N-2) (M+N-1) (M+N)

These factors constitute the 2N+1 consecutive non-negative integers M-N through M+N.

That the product of K consecutive non-negative integers is divisible by K! can be seen by considering combinatorial notation.

Let K and L be any non-negative integers. Consider the number of K-element sets of an (K+L) element set. This is given by the combinatorial expression C(K+L,K), which is always an integer.

Now, C(K+L,K) = (L+1)(L+2)(L+3)...(L+K) / K!.

The numerator is the product of K consecutive integers, and the denominator is K!. Since K and L are arbitrary non-negative integers, the product of K consecutive non-negative integers is always divisible by K!.

[Antares]

Monte, you're an utter genius. I bow in reverence to the Great Mathematician. It makes just an inkling of sense to me, since I'm in Algebra II/Trig, but yes, you're an utter genius.

[Monte314]

Monte, you're an utter genius. I bow in reverence to the Great Mathematician. It makes just an inkling of sense to me, since I'm in Algebra II/Trig, but yes, you're an utter genius.

That is very high praise indeed coming from you.

(Actually, I'd be satisfied with "Nice hair, dude.")

[Sequoia]

Yikes, I'm feeling old. Once upon a time, I would have gotten this easily, but as the decades passed, things have gotten more than a tad rusty. It almost looks like a foreign language now. But not quite.