Divizibilitate

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Mateescu Constantin
Newton
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Divizibilitate

Post by Mateescu Constantin »

Fie \( p \) un numar prim, \( p\ge 3 \). Sa se arate ca numarul \( \frac{(p-1)!}{\left\[\left\(\frac{p-1}{2}\right\)\Large{!} \right\]^2}+(-1)^{\frac{p+1}{2}} \) este divizibil cu \( p \).
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SNT
Euclid
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Joined: Wed Dec 26, 2007 5:11 pm
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Post by SNT »

Se observa imediat ca aducand la acelasi numitor, obtinem ... = (p-1)! - (p-1)! = 0 (mod p), deoarece x = (-1)(p-x) (mod p), pt. x de la 1 la (p-1)/2.
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