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 \((a,b) \in \mathbb{Q}^{2}\).

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## Comments

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TopNewestIt 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.

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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 \(a=b=c=1\).

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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.

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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?

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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.

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