Show that \[ (x^2+2y^2)^2-4x^2y^2=x^4+4y^4, \]

which is what we wanted.

and hence factorise \(x^4+4y^4\).

We know that \[ x^4+4y^4=(x^2+2y^2)^2-4x^2y^2. \] Notice that the right hand side is of the form \(a^2-b^2\), where \(a=x^2+2y^2\) and \(b=2xy\).

We know that \(a^2-b^2=(a+b)(a-b)\), and so \[ (x^2+2y^2)^2-(2xy)^2=(x^2+2y^2+2xy)(x^2+2y^2-2xy). \]

Can we factorise the terms in the brackets any further? We can think of the first bracket as \[
x^2+2yx+2y^2,
\] which is a quadratic in \(x\) with coefficients that depend on \(y\). If this factorised into \((x+ay)(x+by)\) for some \(a\) and \(b\), then there would be one or two real solutions to \(x^2+2yx+2y^2=0\) for each real value of \(y\). So we can check the discriminant,

\[
(2y)^2-4(2y^2)=4y^2-8y^2=-4y^2.
\] But this is less than zero whenever \(y\) is non-zero. Thus \(x^2+2y^2+2xy=0\) doesn’t have any real solutions if \(y\ne0\), and so we can’t factorise the first bracket any further.

For the second bracket, the discriminant is also \[ (-2y)^2-4(2y^2)=-4y^2, \] and so we can’t factorise the second bracket either. Therefore we’ve fully factorised \(x^4+4y^4\).

Alternatively we can see \((x^2+2y^2+2xy)(x^2+2y^2-2xy) = ((x-y)^2+y^2)((x+y)^2+y^2)\), which won’t factorise over the real numbers.

It turns out therefore that \(x^2+4y^2\) does not factorise, but \(x^4+4y^4\) does. Why do they behave so differently?