Ecuatie in numere naturale - JBTST III 2007, problema 2
Posted: Tue Apr 08, 2008 6:14 pm
Sa se rezolve in numere naturale ecuatia
\( (x^{2}+2)(y^{2}+3)(z^{2}+4)=60xyz \).
\( (x^{2}+2)(y^{2}+3)(z^{2}+4)=60xyz \).
Remarcam, mai intai, ca ecuatia \( (x^2+2)(y^2+3)(z^2+4)=60xyz\ (1) \), daca are solutiii in \( \mathbb{N} \), atunci \( x,\ y,\ z\in\mathbb{N}^{\ast} \)
Avem relatiile echivalente: \( \left\{\begin x^2+2\ge 3x\ \Longleftrightarrow (x-1)(x-2)\ge 0,\ \mbox{cu egalitate daca}\ \ x\in\{1,\ 2\}\ (2) \\y^2+3\ge 4y\ \Longleftrightarrow (y-1)(y-3)\ge 0,\ \mbox{cu egalitate daca}\ y\in\{1,\ 3\}\ (3) \\z^2+4\ge 5z\ \Longleftrightarrow (z-1)(z-4)\ge 0,\ \mbox{cu egalitate daca}\ z\in\{1,\ 4\}\ \ \ (4) \)
Din \( (2) \), \( (3) \) si \( (4) \) se observa ca \( (x^2+2)(y^2+3)(z^2+4)\ge 60xyz \), iar daca \( x\in\{1,\ 3\},\ y\in\{1,\ 4\},\ z\in\{1,\ 4\} \) avem egalitate.
Deci ecuatia are solutiile: \( (x,\ y,\ z)\in\{(1,\ 1,\ 1);\ (1,\ 1,\ 4);\ (1,\ 3,\ 1);\ (1,\ 3\ ,4);\ (2,\ 1,\ 1);\ (2,\ 1,\ 4);\ (2,\ 3,\ 4);\ (2,\ 3,\ 1)\} \)
Aratam ca ecuatia \( (1) \) nu mai are si alte solutii.
Tot din \( (2) \), \( (3) \) si \( (4) \) se observa ca \( (x^2+2)(y^2+3)(z^2+4)>60xyz \), daca \( y>3,\ z>4,\ \forall x\in\mathbb{N}^{\ast} \), deci urmeaza ca ecuatia \( (1) \) nu are solutii in \( \mathbb{N} \).
Din \( (1) \) si \( (2) \) \( \Longrightarrow (y^2+3)(z^2+4)\le 20yz\ (5) \)
Daca \( y=1 \), din \( (5)\ \Longrightarrow z^2+4\le 5z\ \Longleftrightarrow (z-1)(z-4)\le 0\ \Longleftrightarrow z\in\{1,\ 2,\ 3,\ 4\} \). Se arata ca \( z\in\{2,\ 3\} \) si \( y=1 \) nu conduc la solutii.
Daca \( y=2 \), din \( (5)\ \Longrightarrow 7z^2-40z+28\le 0\ \Longleftrightarrow z\le 5 \). Verificam in \( (1) \) ca \( z\le 5 \) si \( y=2 \) (nu conduc la solutii).
Daca \( y=3 \), din \( (5)\ \Longrightarrow 12(z^2+4)\le 60z\ \Longleftrightarrow 3z^2-15z+1\le 0\ \Longleftrightarrow 3z(z-5)+1\le 0\ \Longleftrightarrow z\le 4 \). Aratam ca \( y=3 \) si \( z\in\{2,\ 3\} \) nu conduc la solutii.
Daca \( z=1 \), din \( (5)\ \Longrightarrow y^2-4y+3\le 0 \) de unde \( 1\le y\le 3 \), caz analizat.
Daca \( z=2 \) vom avea \( y\in\{2,\ 4,\ 5\} \) care nu dau solutii.
Analog se analizeaza \( z\in\{3,\ 4\} \).