Matematické Fórum

stuart clark · 02. 02. 2016 12:21

The number of all real solution $(x,y)$ of the equation $16x^4+y^4+8x^2-2y^2-24xy+8=0$

check_drummer

df/dx is wrongly computed...

Brano · 04. 02. 2016 11:22

↑ stuart clark:
first take s look here to see that the answer is infinitely many

now to prove it denote $f=16x^4+y^4+8x^2-2y^2-24xy+8$
and observe that
$f(1,2)=-8<0$
$f(0,y)=y^4-2y^2+8=(y^2-1)^2+7>0$
$f(x,0)=16x^4+8x^2+8=(4x^2+1)^2+7>0$
$f(2,y)=y^4-2y^2+48y+296=(y^2-2)^2+2(y-12)^2+4>0$
$f(x,3)=16x^4+8x^2-72x+71=(4x^2-2)^2+6(2x-3)^2+13>0$

Denote by $S$ the boundary of the square with corners $(0,0);\ (0,3);\ (2,3);\ (2,0)$ and $P=(1,2)$ . Now for any segment connecting $P$ with point on $S$ we have at least one solution of $f=0$ , which yields infinitely many solutions.