This theorem belongs to Gao Weidong.

I found it and copied it to everyone.

Firstly, it is proved that it is true for prime number p.

If there is a counterexample A 1A2 ... I; one

Take any p and sum (x 1+x2+...+XP) ≠ 0 (MOPP).

Then according to Fermat's last theorem, (x1+x2+...+XP) (p-1) =1(MOPP).

Running around for peace,

∑ (x1+x2+...+XP) (p-1) = ∑1(MOP Pass

Right =C(2p- 1，p)≡(2p- 1)！ /[(p！ )(p- 1)！ ]≡ 1(mod p)

(Cut off P and use Wilson's theorem)

What is left equal to? One of them (taking p= 13 as an example) is as follows.

A12 * a25 * a34 * a4 (item m, power is divided into K 1, K2, ..., km).

* * * Expand in several (α) brackets, and each bracket provides several (β) items. A * * * is

αβ=C(2p- 1-m, p-m)*β. Then something like C will definitely be divisible by P, won't it? Just write it.

So the ∑ on the left is 0(mod p), which is inconsistent with 1 on the right.

Yes, P will prove it.

Suppose a and b are proved to satisfy singular conditions.

For the number {x_n} of 2AB- 1, a is extracted from it, and the sum of A 1 is a multiple of a; Then a is extracted, and the sum is also required to be a multiple a2 of a; You can get 2B- 1 group at any time. (This leaves a lonely A- 1 in the end.)

Look at A 1, A2, ... A _ (2B- 1). Naturally, we can take out B and make the sum a multiple of B.

The x corresponding to these numbers is the number of AB {x_l}, which is a multiple of a or a multiple of b after summation.

This proves the proposition.

I think I can write a certificate. Two trivial solutions of K=2 and K=N are eliminated.

There is a connection between two points, even a red line; There is no connection, even the blue line.

For a certain (n, k), if there is a counterexample, that is, the small organizations in point K are United, but point N is not, we will examine this counterexample and remove it.

Just find all the k's with counterexamples.

For a counterexample diagram G0, there must be a blue line on each of the n points, but any small organization of the k points is very red and United.

G0 If some blue lines are appropriately deleted and replaced by red lines, the G obtained must be a counterexample.

How to delete it is cleaner?

Points x 1 and x2 are connected with blue lines, and these two points are recorded as a complete B2. If there is no blue line connecting other points with B2, it is said that B2 is an extension of x 1. If x3 and B2 have a blue line, then x 1 x2 x3 is B3; Expand it like this until it can't expand any more. This b contains m points, m- 1 blue line. Call this figure "Road". Are roads like hydrocarbons?

Then choose an expansion center from the remaining points and expand it again to get the expanded path C.

d、E……

I remember that the blue line system was reduced to A 1, A2, ... (in descending order of number of members)

We have to prove that for some K's, there will always be a "blue" K-point group that is not too red and not too United. (that is to say, the counterexample is untenable-one mouth and one brain = =! )

The above is the preparation work. When n is odd, A 1 is not less than 3.

If graph G is the dimension reduction result of G0(N, k) as a counterexample.

K is w = as+as- 1+...+ar-mm! Can't add any more, adding one will exceed the (or arrival) time.

If K & gtW+ 1=W+w, then as+as-1+...+ar draws the first w points in A 1, which are exactly k and each point has a blue line.

If K=W+ 1, there are still counterexamples that can be properly constructed. You can try it yourself. If you understand the spirit of this proof. (If As = 2 ... If not ...)

Explain that there is no counterexample for n-bit odd numbers.

In the same way, for n, it is an even number, and for counterexamples, there are A 1, A2, ..., and so on.

It would be interesting if everyone was 2 years old, wouldn't it? We prove it.

1) if a1>; =3, as discussed in the odd case, there is no such counterexample.

2) If A 1=2

That means x 1x2x3x4x5x6 ... that's right.

At this time, "counterexample" can be easily overturned, just as k is an even number.

It is not difficult to prove that odd k bits are just counterexamples.

To sum up, n is an odd number and k can be taken from 2 to n; N is an even number, and k is an even number between 2 and n.

It should be good.