real solution

[man111]

to find real solution of system of equation:

[tex]\left(x^2-y^2\right)^2 = 1+16y[/tex]

[martin123456]

this is not a system
what is the second equation?

[man111]

martin there is only one equation that is [tex](x^2-y^2)^2 = 1+16y[/tex]

[martin123456]

Solving the equation with [tex]x^2[/tex] as a variable we get [tex]x_{1,2}^2=y^2 \pm \sqrt{8y+1}[/tex].
Case 1: [tex]x^2=y^2+\sqrt{8y+1} \Leftrightarrow x^2-y^2=\sqrt{8y+1} \Rightarrow (x^2-y^2)^2=8y+1[/tex]. So [tex]1+16y=8y+1 \Leftrightarrow y=0 \Rightarrow x=\pm 1[/tex]. Checking [tex](x,y)=(1,0)[/tex] and [tex](x,y)=(-1,0)[/tex] yields that both are solutions.
Case 2: [tex]x^2=y^2-\sqrt{8y+1} \Leftrightarrow x^2-y^2=-\sqrt{8y+1} \Rightarrow (x^2-y^2)=8y+1[/tex], e.g. the same as case 1.

[1089]

(4;3),(5:4) are also solutions...

[martin123456]

yes, i have a mistake
will think out the problem again

[mkmarinov]

By "solving the equation for x" you would get
[tex]x^2=y^2 \pm \sqrt{16y+1}[/tex], which you get by taking the square root of both sides in the original form of the equation.
And the equation obviously has infinitely many solutions, so I guess that it is a mistake in the problem definition (pick a random number y - most of the times there would be a x, for which the equaiton is true).

[1089]

x,y integers, I guess

[man111]

yes [tex]x,y\in\mathbb{Z}[/tex]