# 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
6 years, 7 months ago

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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 - 6 years, 7 months ago

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

- 6 years, 7 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.

- 6 years, 7 months ago

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

Maybe then some casework, and we're done?

- 6 years, 7 months ago

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

- 6 years, 7 months ago

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

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

So $\sqrt{2}$ must be rational. So $\sqrt{2}=\dfrac{p}{q}$ 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.

- 6 years, 7 months ago