Find all real solutions of this equation $x^2=2y-1$,$x^4+y^4=2$. Find all real solutions of this equation $x^2=2y-1$,$x^4+y^4=2$.
My attempt:I put the value of $x^2$ in the second equation.I get:
$(2y-1)^2+y^4=2 \Rightarrow [(2y-1)^2-1^2]+(y^4-1^4)=0 \Rightarrow 4y(y-1)+(y-1)(y+1)(y^2+1)=0 \Rightarrow (y-1)(y^3+y^2+5y+1)=0$
Now one solution is $y=1,x=1or-1$ but what about others is there another real solution?

There are no other real solutions.
From $2y-1=x^2\ge 0,$ $y$ has to satisfy $y\ge \frac 12$.
However, if $y\gt 0$, then $y^3+y^2+5y+1\gt 0$.
