Ecuatie diofantica
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Ecuatie diofantica
Sa se gaseasca toate perechile \( (x,y) \) de numere intregi cu proprietatea : \( x^3-4xy+y^3=-1 \) .
. A snake that slithers on the ground can only dream of flying through the air.
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maxim bogdan
- Thales
- Posts: 106
- Joined: Tue Aug 19, 2008 1:56 pm
- Location: Botosani
Solutie
\( x^3-4xy+y^3=-1\Longleftrightarrow x^3+y^3+1-4xy=0\Longleftrightarrow x^3+y^3+(\frac{4}{3})^3-3\cdot xy\cdot\frac{4}{3}=\frac{37}{27}\Longleftrightarrow (3x)^3+(3y)^3+4^3+3\cdot (3x)\cdot (3y)\cdot 4=37. \)
Aplicand cunoscuta identitate: \( a^3+b^3+c^3-3abc=\frac{1}{2}\cdot (a+b+c)[(a-b)^2+(b-c)^2+(c-a)^2], (\forall) a,b,c\in\mathbb{R} \) obtinem:
\( (3x+3y+4)[(3x-4)^2+(3y-4)^2+(3x-3y)^2]=74 \)
De aici e usor de rezolvat analizand toate cazurile.
Aplicand cunoscuta identitate: \( a^3+b^3+c^3-3abc=\frac{1}{2}\cdot (a+b+c)[(a-b)^2+(b-c)^2+(c-a)^2], (\forall) a,b,c\in\mathbb{R} \) obtinem:
\( (3x+3y+4)[(3x-4)^2+(3y-4)^2+(3x-3y)^2]=74 \)
De aici e usor de rezolvat analizand toate cazurile.
Feuerbach