Conor McBride asks for a fast proof that $$x = {\left(\pi^\pi\right)}^{\pi^\pi}$$ is not an integer. It would be sufficient to calculate a very rough approximation, to a precision of less than $1,$ and show that $n < x < n+1$ for some integer $n$. But $x$ is large enough ($56$ digits) that that method is not obviously practical without electronic computation tools.

Is there a purely theoretic proof that works with no calculation at all, or is there a calculational argument that would be practical prior to $1950$?

Edit: Let me phrase the question a little differently, as a spur to your mathematical imagination. I go back in time to $1847$ and have lunch with Carl Friedrich Gauss. We are chatting about the developments in $20$th and $21$st-century mathematics, and because I am a horrible person, I mention that $(\pi^\pi)^{\pi^\pi}$ has been proven to be an integer. Gauss is incredulous, but I insist. After I leave, can Gauss convince himself that I am nothing but a troll?

Edit: After I looked back at McBride’s original question, I saw to my dismay that he had not asked about $x= (\pi^\pi)^{\pi^\pi}$, but $$y=\pi^{\left(\pi^{\pi^\pi}\right)}$$ which is a rather different animal. The value of $x$ has $57$ decimal digits; the value of $y$ has $10^{57}$digits. So Alexander Walker’s impressive display of direct numerical calculation will avail nothing for deciding that $y$ is not an integer, neither in $1873$ nor in $2013,$ and perhaps not in $2153$ either. (“We should attempt to destroy the aliens.”) More powerful theoretical methods will be required. I am posting an additional bounty to see if anyone can suggest anything of that sort; it need not be theory that was available in the $19$th century.

**Answer**

I think this calculation would have been doable in 1873, as follows:

(1) Compute $\log \pi$ to $60$ digits. For this, we begin with the expansion

$$\log \pi = \log \left(\frac{\pi}{22/7}\right)+\log 2 + \log 11-\log 7$$

and take full advantage of the fact that the logarithm tables of small integers were known to great precision. As for the logarithm of $\pi/(22/7) \approx 1.00041$, the Taylor series for the logarithm spits out $60$ digits after only $17$ terms. (Surprisingly, this exact analysis has been done on MSE before.)

(2) Compute $(\pi+1)\log \pi$ to $60$ digits. This is no big deal, given our value for $\log \pi$ from (1), since $\pi$ was known (to Gauss, no less!) to at least $100$ digits back in 1844. For reference, this value is

$$ \approx 4.7410048855785583722294291029994190930164741026691888020108672.$$

(The multiplication will of course be a pain, as it requires around $1800$ flops. Nevertheless, this computation would likely be delegated to a lesser mathematician. The Wikipedia article on Zacharias Dase (a temporary assistant to Gauss) puts these computations into perspective:

At age 15 [Zacharias Dase] began to travel extensively, giving exhibitions in Germany, Austria and England. Among his most impressive feats, he multiplied 79532853 × 93758479 in 54 seconds. He multiplied two 20-digit numbers in 6 minutes; two 40-digit numbers in 40 minutes; and two 100-digit numbers in 8 hours 45 minutes. The famous mathematician Carl Friedrich Gauss commented that someone skilled in calculation could have done the 100-digit calculation in about half that time with pencil and paper.

(3) Exponentiate the product, again to $60$ places. Using just the series

$$e^z = \sum_{k=0}^\infty \frac{z^k}{k!}$$

such a calculation would require $77$ terms. For this reason, we instead calculate

$$e^{4.741} \approx 114.54869311801681310751748724665811195370661075419665168411647;$$

$$e^{0.0000048\cdots} \approx 1.0000048855904928305900123833767696556988185632721564706179420.$$

The latter approximation requires a mere $10$ terms of the exponential Taylor series to achieve $60$ digits, a commitment of around 18000 flops. By Gauss’s metric, we might expect a skilled mathematician to complete this task in just over seven hours.

The prior calculation could be done in one of two ways: directly, at a cost of another $18000$ flops (e.g. $77$ consecutive multiplications of a $4$-digit and a $60$-digit number); or by calculating $\mathrm{exp}(4)$ and $\mathrm{exp}(.741)$ independently (a slight time savings, I believe, even after the final multiplication). Of course, now it just takes another $1800$ flops to multiply these numbers to $60$ places.

*Note: In hindsight, it appears that calculating the triple product*

$$e^4 e^{3/4}e^{-9/1000}$$

*may expedite this most recent step.*

In case you’ve lost track, we now know the value of

$$e^{(\pi +1)\log \pi}=\pi^{\pi+1}$$

to $60$ digits, all within a day or two of starting our calculations.

(4) Multiply $\log \pi$ and $\pi^{\pi+1}$ to $60$ digits. This step is easy, given steps (1) and (3). This value is

$$\approx 131.12795303153615589452803943707399841542170349230159549341360.$$

Of course, this value is *also* the logarithm of $(\pi^\pi)^{\pi^\pi}$, so it remains to:

(5) Exponentiate the term from (4). Since it worked out so well in (3), we’ll again split our exponential into a product of simpler terms. Here, we luck out – since the binary expansion of $131.127953\ldots$ begins as

$$10000011.001000001100000\cdots_2,$$

the partial approximation $131+1/8$ is surprisingly accurate: to within $\approx 0.002953$. The exponential of this remainder can be made accurate to over $60$ digits with a mere $18$ terms (i.e. $32000$ flops).

Secondly, we compute $e^{131}$ to $60$ digits, using iterated multiplication and an approximation of $e$ to $62$ digits ($205$ digits were known to William Shanks in 1871). Since it suffices here to successively compute

$$e^2,e^4,e^8,e^{16},e^{32},e^{64},e^{128},$$

this step can be done in less than $15000$ flops. Thirdly, we compute $e^{1/8}$ to $60$ digits using three applications of Newton’s method for the square root (another $6000$ flops). We find a value of

$$ \approx 887455172183124295874631455225434602688412866765466125005\color{red}{.16},$$

a non-integer.

All said and done, I would be surprised if this calculation took much longer than a week. At the same time, I stress that this problem would have been *barely* doable in that era. If twice the number of digits were required, for example, this work may very well have taken the better part of a year.

**Attribution***Source : Link , Question Author : MJD , Answer Author : Community*