Is $5^\pi$ an integer?

Question

I saw this very short math problem on Twitter:

Is $5^\pi$ an integer?

It isn't (it's 156.992545309…), but is there some technique to prove this without a calculator?

My first and only idea so far is to prove $156<5^\pi<157$ by taking logarithms, but it's annoyingly close to 157, and anyway I used a calculator to find those bounds.

Answer 1

Proving the result by hand may take some effort, but the following process shows that it can be done. In what follows, one assumes the known estimates for $\pi$ and $e:$

$$\frac{201}{64}<\frac{333}{106}<\pi<\frac{355}{113}<\frac{3217}{1024}\qquad (1a)$$ and $$2.7182<\frac{1264}{465}<e<\frac{1457}{536}<2.7183.\qquad (1b)$$ By (1a), one has $$5^{\frac{201}{64}}<5^{\pi}<5^{\frac{3217}{1024}}.$$ To show that $5^{\pi}$ is not an integer, it suffices to show that $$156<5^{\frac{201}{64}}~{\rm and}~5^{\frac{3217}{1024}}<157.\qquad (2)$$ One completes the proof of (2) via the following Lemmas, where Lemma 1 is a direct consequence of the Alternating Series Test.

Lemma 1. Let $0<x\leq 1.$ Then $$\sum_{k=1}^{2n}(-1)^{k-1}\frac{x^k}{k}<\ln(1+x)=\sum_{k=1}^{\infty} (-1)^{k-1}\frac{x^k}{k}<\sum_{k=1}^{2n-1}(-1)^{k-1}\frac{x^k}k,\forall n\in{\mathbb N}.$$

Lemma 2. The inequalities $1.6094<\ln 5<1.60951$ hold.

Proof. One has $$\ln 5=\frac 3 2+\frac 12\ln\left(\frac{25}{e^3}\right).$$ It follows from (1b) that $$1.2446<\frac{25}{2.7183^3}<\frac{25}{e^3}<\frac{25}{2.7182^3}<1.2448.$$ Using Lemma 1 with $x=0.2448,$ one has $$\ln 5<\frac 3 2+\frac 1 2\left(x-\frac{x^2}2+\cdots+\frac{x^5}5\right)<1.60951.$$ Similarly with $x=0.2446,$ one has $$\ln 5>\frac 3 2+\frac 1 2\left(x-\frac{x^2}2+\cdots-\frac{x^6}6\right)>1.6094.$$ Hence the result follows.

Lemma 3. The inequality $156<5^{\frac{201}{64}}$ holds.

Proof. This is equivalent to showing that $$\ln(156)<\frac{201}{64}\ln 5$$ $$\Leftrightarrow \ln\frac{156}{125}+3\ln 5<\frac{201}{64}\ln 5$$ $$\Leftrightarrow \ln\left(1+\frac{31}{125}\right)<\frac 9{64}\ln 5.$$ Now by Lemma 1, one has $$\ln\left(1+\frac{31}{125}\right)<x-\frac{x^2}2+\frac{x^3}3\sim 0.22233,$$ where $x=\frac{31}{125}.$ By Lemma 2, one has $$0.22632\sim \frac 9{64}(1.6094)<\frac 9{64}\ln 5.$$ Hence the result follows.

Lemma 4. The inequality $5^{\frac{3217}{1024}}<157$ holds.

Proof. This is equivalent to $$\frac{3217}{1024}\ln 5<\ln(157)$$ $$\Leftrightarrow \frac{3217}{1024}\ln 5<3\ln 5+\ln\frac{157}{125}$$ $$\Leftrightarrow \frac{145}{1024}\ln 5<\ln\left(1+\frac{32}{125}\right).$$ Now by Lemma 2, one has $$\frac{145}{1024}\ln 5<\frac{145}{1024}(1.60951)\sim 0.22791.$$ By Lemma 1, one has $$\ln\left(1+\frac{32}{125}\right)>x-\frac{x^2}2+\cdots+\frac{x^5}5-\frac{x^6}6\sim 0.22792,$$ where $x=\frac{32}{125}.$ Hence the result follows.

Combining Lemma 3 and Lemma 4, the result is proven. QED

Answer 2

First note: I am not a mathematician, nor I have any mathematical training besides school, so my words should be taken with a grain of salt, a mountain of salt.

We can see by the approximation that (1) $3.1414 < \pi < 3.1416$ and, by asking Wolfram|Alpha, that (2) $5^{31416} < 157^{10'000}$ and (3) $5^{31414} > 156^{10'000}$. As you can calculate all of these by hand, I will assume it is acceptable to use those sources to just accelerate the process, as I have done. We now prepare the range-value of pi,

$3.1414 < \pi < 3.1416 $

$5^{3.1414} < 5^{\pi} < 5^{3.1416} $

$5^{3.1414} < 5^{\pi} \text{ and } 5^{\pi} < 5^{3.1416} $

(4) $5^{\pi} > 5^{3.1414} \text{ and } 5^{\pi} < 5^{3.1416} $

Now, we prepare inequalities (2) and (3),

$5^{31414} > 156^{10'000}$ and $5^{31416} < 157^{10'000}$

$5^{\frac{31414}{10'000}} > 156$ and $5^{\frac{31416}{10'000}} < 157$

(5) $5^{3.1414} > 156$ and $5^{3.1416} < 157$

We now use facts (4) and (5):

$5^\pi > 156$ and $5^\pi < 157 $

$156 < 5^\pi$ and $5^\pi < 157 $

$156 < 5^\pi < 157 $

As there is no integer between $156$ and $157$, then $5^\pi$ is not an integer.

Answer 3

This proof is intended be a more elementary than the existing proofs, relying more on "brute force". In exchange, it's more tedious, as you need to manually compute a bunch of square roots and multiplications.

As input, I assume that one knows that: $\pi$ has an expansion which is about 3.14159265...

We will try to prove that $5^\pi$ is not an integer by multiplying $5^3 \times 5^\frac{1}{10} \times 5^\frac{4}{100} \times \ldots$ and using bounds.

Immediately, the problem one runs into is: how do you compute powers like $5^\frac{4}{100}$ by hand? That's not usually covered in school math, but one does learn how to compute square roots by hand. So let's work with the expansion in base 2 instead of base 10.

$$ 0.14159265... (\textrm{base 10}) \rightarrow 0.00100100001... (\textrm{base 2}) $$

We should first compute a bunch of values for $5^\frac{1}{2^N}$. A good "checkpoint" value for $N$ that we will need (roughly) is such that $5^\frac{1}{2^N} < 1.008$, because we will end up wanting to bound:

$$ 5^{3 + \frac{1}{8} + \ldots} < 5^\pi < 5^{3 + \frac{1}{8} + \ldots + \frac{1}{2^N}} $$

and $5^3 \times 0.008 = 1$.

Tediously computing a bunch of square roots by hand:

$$ 5^\frac{1}{2} = 2.23606\ldots \\ 5^\frac{1}{4} = 1.49534\ldots \\ 5^\frac{1}{8} = 1.22284\ldots \\ 5^\frac{1}{16} = 1.10582\ldots \\ 5^\frac{1}{32} = 1.05158\ldots \\ 5^\frac{1}{64} = 1.02546\ldots \\ 5^\frac{1}{128} = 1.01265\ldots \\ 5^\frac{1}{256} = 1.00630\ldots $$

Let's try the following:

$$ 5^{3 + \frac{1}{8} + \frac{1}{64}} < 5^\pi < 5^{3 + \frac{1}{8} + \frac{1}{64} + \frac{1}{256}} $$

For the left side:

$$ 125 \times 1.2228 \times 1.0254 < 5^{3 + \frac{1}{8} + \frac{1}{64}} < 5^\pi \\ 156.73\ldots < 5^\pi $$

For the right side:

$$ 5^\pi < 5^{3 + \frac{1}{8} + \frac{1}{64} + \frac{1}{256}} < 125 \times 1.2229 \times 1.0255 \times 1.0063 \\ 5^\pi < 157.74\ldots $$

Drat! We still have one integer $157$ within the bounds. But this was expected, because the interval size we're working with is roughly 1, so when we got a lower bound of 156.73, the upper bound should be around 157.73.

At this, point, the next 1 in the binary expansion of $\pi$ is 3 more digits away, so the value must be relatively close to the lower bound. Let's try constraining the upper bound more.

$$ 5^\frac{1}{512} = 1.0031\ldots \\ 5^\frac{1}{1024} = 1.00157\ldots $$

Retrying:

$$ 5^\pi < 5^{3 + \frac{1}{8} + \frac{1}{64} + \frac{1}{1024}} < 125 \times 1.22285 \times 1.02547 \times 1.00158 \\ 5^\pi < 156.99... $$

So $156 < 5^\pi < 157$.