An open problem

My friend posed this question:

Does there exist an ordered pair $(a,b) \in \mathbb{Z}^{2}$ satisfying $a+b\pi = \pi^{\pi}$ ?

So far I've tried looking at properties of $\lbrace x\pi \rbrace-\lbrace \pi^{\pi} \rbrace$ , at a few series representations of $\pi$ , and at Mahler's classification of transcendental numbers (which I can't even begin to comprehend).

I hope someone more versed in linear algebra can provide a proof of $a+b\pi \neq \pi^{\pi}$ , extending to $(a,b) \in \mathbb{Q}^{2}$ .

#Algebra

Easy Math Editor

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Comments

It is not known if $\pi ^{ \pi }$ is rational, algebraic or transcendental.

If such an ordered pair exists, then $\pi ^ { \pi }$ must be transcendental (since it is clear that $b \neq 0$ ).

Hence, there currently does not exist a proof if such an ordered pair exists.

Calvin Lin Staff - 6 years, 6 months ago

Kind of related: Prove that if $a+b\sqrt[3]{2}+c\sqrt[3]{4}=1+\sqrt[3]{2}+\sqrt[3]{4}$ for $a,b,c\in \mathbb{Z}$ , prove that $a=b=c=1$ .

Daniel Liu - 6 years, 6 months ago

When an integer $x$ is not a perfect cube, the vectors $x^{0}$ , $x^{1}$ and $x^{2}$ are linearly independent? I don't know how one would go about proving that.

Jake Lai - 6 years, 6 months ago

An idea is substituting $a'=a-1$ , $b'=b-1$ , and $c'=c-1$ . Then you just need to prove that given $a'+b'\sqrt[3]{2}+c'\sqrt[3]{4}=0$ , prove that $a'=b'=c'=0$ which seems simpler to me.

Maybe then some casework, and we're done?

Daniel Liu - 6 years, 6 months ago

@Daniel Liu – Then you can rewrite as $b'\sqrt[3]{2}+c'\sqrt[3]{4}=-a'$ and cubing both sides, rearranging gives $-6b'c'\left(\sqrt[3]{2}+\sqrt[3]{4}\right)=a'^3+2b'^3+4c'^3$ . Straightforward now.

Jubayer Nirjhor - 6 years, 6 months ago

@Jubayer Nirjhor – So we just need to prove that $\sqrt[3]{2}+\sqrt[3]{4}$ is irrational because $-\dfrac{a'^3+2b'^3+4c'^3}{6b'c'}$ is rational.

Suppose that $\sqrt[3]{2}+\sqrt[3]{4}$ is rational. Then $\dfrac{1}{1+\sqrt[3]{2}+\sqrt[3]{4}}=\sqrt[3]{2}-1$ is also rational.

So $\sqrt[3]{2}$ must be rational. So $\sqrt[3]{2}=\dfrac{p}{q}$ for coprime positive integers $p,q$ .

Cubing both sides, $2=\dfrac{p^3}{q^3}$ so $p^3=2q^3$ so $p$ is even.

Let $p=2p'$ so $8p'^3=2q^3$ so $4p'^3=q^3$ and $q$ is even.

But this contradicts our original assumption that $p,q$ are relatively prime so we are done.

Daniel Liu - 6 years, 6 months ago

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Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
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`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$