Find all the values of parameter a, when system has 2 solutions.

Question

x + y = a,

x^4 + y^4 = a.

At first the system was a little bit more difficult, but I simplified it, and got this system. I've tried graphical solution, but I've got only the answer with no ideas, how to explain it.

Answer 1

Hint: Use that $$x^4+y^4+4xy(x^2+y^2)+6x^2y^2=a^4$$ and now use $$x^2+y^2=a^2-2xy$$ so you will get $$a+4xy(a^2-2xy)+6x^2y^2=a^4$$ form here you will get $$xy$$

Answer 2

Write the second as:

$$ x^4+4\left(\frac{y}{\sqrt{2}}\right)^4=a $$

By Sophie Germain Identity:

$$ (x^2+y^2+\sqrt{2}xy)(x^2+y^2-\sqrt{2}xy)=a $$

Notice that:

$$x^2+y^2=(x+y)^2-2xy=a^2-2xy$$

This implies:

$$(a^2-2xy+\sqrt{2}xy)(a^2-2xy-\sqrt{2}xy)=a$$

Now if $X=xy$, you can solve for $X$ and then it should be easy

:)