×

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

Note by Jake Lai
2 years, 3 months ago

Sort by:

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. Staff · 2 years, 3 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 $$a=b=c=1$$. · 2 years, 3 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. · 2 years, 3 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? · 2 years, 3 months ago

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. · 2 years, 3 months ago

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. · 2 years, 3 months ago