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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?

aradarbel10
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    Well, you found an example of $a=b=0$ where neither is irrational, didn't you? Perhaps you should rephrase the question to ask about "For positive" rather than "For real" – JMoravitz Dec 11 '19 at 20:08
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    You can't, since both $a,b$ may be equal to $0$ and neither is irrational. Do you have to assume that they are nonzero. To answer your more specific question, it is invalid to plug one equation into the other, since all you can deduce is that one of those equalities holds (with one of the sign choices) – Wojowu Dec 11 '19 at 20:09
  • @JMoravitz Thanks, edited question – aradarbel10 Dec 11 '19 at 20:13
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    If $a$ and $b\ne 0$ were both rational , then $\frac{a^2}{b}$ would be rational as well, which is impossible because it is $\pm \sqrt{2}$ – Peter Dec 11 '19 at 20:17

2 Answers2

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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.

kccu
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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.

Ty Jensen
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