An open problem

My friend posed this question:

Does there exist an ordered pair (a,b)Z2(a,b) \in \mathbb{Z}^{2} satisfying a+bπ=ππa+b\pi = \pi^{\pi}?

So far I've tried looking at properties of {xπ}{ππ}\lbrace x\pi \rbrace-\lbrace \pi^{\pi} \rbrace, at a few series representations of π\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+bπππa+b\pi \neq \pi^{\pi}, extending to (a,b)Q2(a,b) \in \mathbb{Q}^{2}.

Note by Jake Lai
6 years, 7 months ago

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

If such an ordered pair exists, then ππ \pi ^ { \pi } must be transcendental (since it is clear that b0 b \neq 0 ).

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

Calvin Lin Staff - 6 years, 7 months ago

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Kind of related: Prove that if a+b23+c43=1+23+43a+b\sqrt[3]{2}+c\sqrt[3]{4}=1+\sqrt[3]{2}+\sqrt[3]{4} for a,b,cZa,b,c\in \mathbb{Z}, prove that a=b=c=1a=b=c=1.

Daniel Liu - 6 years, 7 months ago

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When an integer xx is not a perfect cube, the vectors x0x^{0}, x1x^{1} and x2x^{2} are linearly independent? I don't know how one would go about proving that.

Jake Lai - 6 years, 7 months ago

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An idea is substituting a=a1a'=a-1, b=b1b'=b-1, and c=c1c'=c-1. Then you just need to prove that given a+b23+c43=0a'+b'\sqrt[3]{2}+c'\sqrt[3]{4}=0, prove that a=b=c=0a'=b'=c'=0 which seems simpler to me.

Maybe then some casework, and we're done?

Daniel Liu - 6 years, 7 months ago

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@Daniel Liu Then you can rewrite as b23+c43=ab'\sqrt[3]{2}+c'\sqrt[3]{4}=-a' and cubing both sides, rearranging gives 6bc(23+43)=a3+2b3+4c3-6b'c'\left(\sqrt[3]{2}+\sqrt[3]{4}\right)=a'^3+2b'^3+4c'^3. Straightforward now.

Jubayer Nirjhor - 6 years, 7 months ago

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@Jubayer Nirjhor So we just need to prove that 23+43\sqrt[3]{2}+\sqrt[3]{4} is irrational because a3+2b3+4c36bc-\dfrac{a'^3+2b'^3+4c'^3}{6b'c'} is rational.

Suppose that 23+43\sqrt[3]{2}+\sqrt[3]{4} is rational. Then 11+23+43=231\dfrac{1}{1+\sqrt[3]{2}+\sqrt[3]{4}}=\sqrt[3]{2}-1 is also rational.

So 23\sqrt[3]{2} must be rational. So 23=pq\sqrt[3]{2}=\dfrac{p}{q} for coprime positive integers p,qp,q.

Cubing both sides, 2=p3q32=\dfrac{p^3}{q^3} so p3=2q3p^3=2q^3 so pp is even.

Let p=2pp=2p' so 8p3=2q38p'^3=2q^3 so 4p3=q34p'^3=q^3 and qq is even.

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

Daniel Liu - 6 years, 7 months ago

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