Proof of the lucas theorem and proof of the lucas Theorem

Source: Internet
Author: User

Proof of the lucas theorem and proof of the lucas Theorem

Http://baike.baidu.com/link? Url = jJgkOWPSRMobN7Zk4kIrQAri8m0APxcxP9d-C6qSkIuembQekeRwUoEoBd6bwdidmoCRQB_dBklDffpzM_87iSPMyiph2iAXCTyv19YpuuG

Let's take a look at Feng Zhigang's proof of elementary number theory.

Final supplement

The form of each item in the (1 + x) a0 power expansion can be written into the b0 power of C (a0, b0) x. Each item is added.
Likewise
(1 + xp) the form of each item in the a1 power expansion can be written into the b1 power of C (a1, b1) (P power of x. Each item is added.
....
Because (1 + x) a0 and (1 + x p to the power of) a1 to the power is multiplied form. So it can be abstracted as a polynomial (x1 + x2 + ..) * (y1 + y2 + y3 ++ ...) * (z1 + z2 + z3 + ....) *... after removing the parentheses, each item has x, y, and z. That is, the form of each item is (x * y * z *....) + ....
The method above is used to write the right side of the equality with the remainder to the power b1 of the b0 power * C (a1, b1) (P power of x) of C (a0, b0) x.
.... It can be converted into C (a0, b0) * C (a1, b1 )*... * x (b0 + b1 * P + b2 * P's square + b3 * p's 3rd power ....), the x index is in the p-base format of B because b0, b1, b2... is unique, so there is a unique sequence so that the above f (x) with the remainder g (x) then f (ai) with the remainder g (bi) so the same remainder is B of C (a, B) x on the left and C (a0, b0) * C (a1, b1) * on the right of the same remainder )*..... * x (B = b0 + b1 * P + b2 * P's square + b3 * p's 3rd power ....) the power coefficient should be the same, that is, the conclusion is true.

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