First we can verify 2003 is a prime.
Then, by Wilson's theorem, we have (2002)!≡-1(mod 2003).
Also, using Wilson's theorem we have the following congruences.
(1^2)(3^2)***(2001^2)≡(-1)^[(2003+1)/2]≡1(mod 2003)
(2^2)(4^2)***(2002^2)≡(-1)^[(2003+1)/2]≡1(mod 2003)
Let 1*3***2001=a and let 2*4***2002=b.
Then I^2≡a^2+b^2+2ab≡1+1+2*(-1)≡0(mod 2003).
Since 2003 is a prime, we obtain I≡0(mod 2003).
Therefore 2003 divides I.
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◆ From: 140.114.222.127
※ 編輯: nwn 來自: 140.114.222.127 (10/02 02:07)