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