For real positive $a,b$ given $a^2b^2(a^2b^2+4)=2(a^6+b^6)$, show that at least one of the numbers is irrational I think I am close to the answer, but still not sure how to finish it.
$$a^2b^2(a^2b^2+4)=2(a^6+b^6)$$
I know I can rewrite that equation as
$$(a^2-b\sqrt2)(a^2+b\sqrt2)(a\sqrt2-b^2)(a\sqrt2+b^2)=0$$
Thus $$a^2=\pm b\sqrt2 \space\space\space\space\space\space b^2=\pm a\sqrt2$$
Now, am I allowed to substitute one equation into the other? This would mean that $b=0$ or $b=\pm\frac{\sqrt[3]4}{\sqrt[3]{\sqrt2}}$. But then when I try $b=0$ I get $a=0$ as well.
How can I prove at least one of them is irrational?
 A: As Wojowu pointed out, you can't plug one equation into another since you only know at least one of them holds. However, you can deduce from both equations individually that $a$ and $b$ cannot both be rational, unless they are both zero.
Suppose the first equation holds: $a^2 = \pm b\sqrt{2}$. Then $a^4=2b^2$, so $2=\frac{a^4}{b^2}$ (assuming $b \neq 0$). Why is this impossible if $a$ and $b$ are both rational? Do a similar argument for the other equation.
A: Lets do it the long way. Distribute everything and move it to one side
$$a^4b^4+4a^2b^2-2a^6-2b^6=0$$
Lets substitute 
$$x=ab$$
Note x is always positive by the positive definition of $1$ and $b$. Then,
$$a=\frac{x}{b}$$
$$b=\frac{x}{a}$$
Which are also positive definite relations. Now substitute to get
$$x^4 + 4x^2 -2(\frac{x^6}{a^6})-2(\frac{x^6}{b^6})=0$$
Factor out an $x^2$
$$x^2(x^2+4-x^4(\frac{2}{a^6} + \frac{2}{a^6}))=0$$
Note that $(\frac{2}{a^6} + \frac{2}{a^6})$ is also positive definite, so lets replace it with another positive definite substitution $c = \frac{2}{a^6} + \frac{2}{a^6}$. Now we have
$$x^2(-cx^6+x^2+4)=0$$
Lets find what the roots of the inside will be, since $x^2=0$ is trivial.
$$-cx^4+x^2+4=0$$
Complete the square
$$x^4-\frac{1}{c}x^2=\frac{4}{c}$$
$$x^4-\frac{1}{c}x^2 +\frac{1}{4c^2}=\frac{4}{c}+\frac{1}{4c^2}$$
$$(x^2-\frac{1}{2c})^2=\frac{16c+1}{4c^2}$$
$$x^2-\frac{1}{2c}=\sqrt{\frac{16c+1}{4c^2}} = \pm \frac{\sqrt{16c+1}}{2c}$$
Finally,
$$x^2=\frac{1}{2c}\pm \frac{\sqrt{16c+1}}{2c}$$
and since everything is positive definite, note that
$$\frac{1}{2c} - \frac{\sqrt{16c+1}}{2c} < 0$$
so we must have at least 1 complex root.
