Concursul "Ion Ciolac" problema 1

BogdanCNFB · Post by **BogdanCNFB** » Sat Apr 25, 2009 12:36 pm

Rezolvati in multimea numerelor intregi ecuatia:
\( x^3+y^3-3xy-3=0 \).

Mateescu Constantin · Post by **Mateescu Constantin** » Sat Apr 25, 2009 4:07 pm

Notam \( x+y=s \), \( xy=p \). Astfel, ecuatia din enunt devine:

\( s(s^2-3p)=3p+3\ \Longleftrightarrow\ s^3-3ps-3p-3=0\ \Longleftrightarrow\ s^3+1-3p(s+1)=4 \)

\( \Longleftrightarrow (s+1)(s^2-s+1)-3p(s+1)=4\ \Longleftrightarrow\ (s+1)(s^2-s+1-3p)=4 \)

Ecuatia fiind in numere intregi il scriem pe 4 ca toate produsele de numere intregi si se calculeaza solutiile in fiecare caz.

Marius Mainea · Post by **Marius Mainea** » Sat Apr 25, 2009 6:19 pm

Folosim formula: \( a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca) \).
Asadar \( x^3+y^3+1^3-3xy\cdot 1=(x+y+1)(x^2+y^2+1-xy-x-y)=4 \).