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Prove Fermat's theorem by expanding via the ... - Cs.ioc.ee

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Urmas Repinski<br />

Tallinn University of Technology<br />

Department of Computer Engin<strong>ee</strong>ring<br />

<strong>Prove</strong> <strong>Fermat's</strong> <strong><strong>the</strong>orem</strong> <strong>by</strong><br />

<strong>expanding</strong> <strong>via</strong> <strong>the</strong><br />

multinomial <strong><strong>the</strong>orem</strong><br />

Concrete Ma<strong>the</strong>matics<br />

Homework 2<br />

2012-10-31


Task definition<br />

<strong>Prove</strong> <strong>Fermat's</strong> <strong><strong>the</strong>orem</strong> n p ≡n(mod p) <strong>by</strong><br />

<strong>expanding</strong> (1+1+...+1) p<br />

<strong>via</strong> <strong>the</strong> multinomial<br />

<strong><strong>the</strong>orem</strong>.<br />

2


Proof<br />

Multinomial Theorem:<br />

(x 1<br />

+x 2<br />

+...+x m<br />

) n = ∑<br />

k 1<br />

+k 2<br />

+...+k m<br />

=n(<br />

n<br />

k 1<br />

, k 2<br />

,... , k m) ∏<br />

1⩽t ⩽m<br />

x t<br />

k t<br />

where<br />

(<br />

n<br />

m) k 1<br />

, k 2<br />

,... , k = n!<br />

k 1<br />

!⋅k 2<br />

!⋅...⋅k m<br />

!<br />

Every sum coefficient is divisible <strong>by</strong> n except if k j<br />

=n,<br />

1≤j≤m. Therefore<br />

(x 1<br />

+x 2<br />

+...+x n<br />

) p ≡x 1 p +x 2 p +...+x n p (mod p)<br />

Setting all x-s to 1 we get<br />

n p ≡n(mod p)<br />

<strong>Prove</strong>n.<br />

3


Multinomial Theorem Informal proof<br />

Multinomial Theorem can be proven <strong>by</strong><br />

induction, and<br />

Coefficients are actually number of distinct ways<br />

how to permute a multiset of n elements, and k i<br />

are multiplicities of each of <strong>the</strong> distinct element.<br />

For example, number of distinct permutations of<br />

letters of <strong>the</strong> world MISSISSIPPI, which has 1M,<br />

4I-s, 4S-s and 2P-s is:<br />

(<br />

11<br />

1,4, 4, 2) = 11!<br />

1!⋅4!⋅4!⋅2! =34650<br />

That’s why it is practically impossible to solve<br />

permutation tasks using random permutations.<br />

4


n p ≡n(mod p)<br />

Thank You<br />

5

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