# Why an almost integer?

Over on https://mathlesstraveled.com there’s a series of posts going on having to do with this:

$\displaystyle \begin{array}{cc} n & (1 + \sqrt 2)^n \\ \hline 1 & 2.414213562373095 \\ 2 & 5.82842712474619 \\ 3 & 14.071067811865474 \\ 4 & 33.970562748477136 \\ 5 & 82.01219330881975 \\ 6 & 197.99494936611663 \\ 7 & 478.002092041053 \\ 8 & 1153.9991334482227 \\ 9 & 2786.0003589374983 \\ 10 & 6725.9998513232185 \end{array}$

See that? As $n$ increases, $(1+\sqrt 2)^n$ approaches integer values. Odd, huh? Why does it do that?

Despite what should have been a dead giveaway hint, I didn’t figure out the approach revealed in this post. Embarrassing. But having failed on the insight front, what can I do on the obvious generalization front?

Let’s think about quantities of the form

$\displaystyle (j+k^{l/m})^n$

where $j$, $k$, $l$, $m$, and $n$ are nonzero integers; $l/m$ is in lowest terms and $l>0$, $m>1$, and $n>0$. For now let’s also restrict $m$ to primes.

To investigate that we’ll consider

$\displaystyle \sum_{p=0}^{m-1} (j+e^{i2\pi p/m}k^{l/m})^n$

The complex quantities $e^{i2\pi p/m}$ lie on the unit circle in the complex plane and are the vertices of an $m$-gon. Using the binomial expansion, the sum is

$\displaystyle \sum_{p=0}^{m-1} \sum_{q=0}^{n} \binom{n}{q} j^{n-q} \left (e^{i2\pi p/m}k^{l/m} \right)^q$

or

$\displaystyle \sum_{q=0}^{n} \binom{n}{q} j^{n-q} \left (\sum_{p=0}^{m-1} e^{i2\pi pq/m}\right) k^{lq/m}$

Now, for the terms where $q$ is a multiple of $m$, $e^{i2\pi pq/m}$ is equal to 1 and the sum over $p$ equals $m$.

Otherwise, we’re summing over the points on the unit circle:

$\displaystyle \sum_{p=0}^{m-1} e^{i2\pi p/m}$

which is the sum of a geometric series so

$\displaystyle = \frac{1-e^{i2\pi}}{1-e^{i2\pi/m}} = 0$

For instance, when $m=2$, the sum is $e^0+e^{i\pi} = 1 - 1 = 0$. When $m = 3$, it’s $e^0 + e^{2i\pi/3} + e^{4i\pi/3} = 1 - \frac{1}{2} - \frac{1}{2} = 0$.

All right then. This means we keep only the terms where $q$ is a multiple of $m$:

$\displaystyle \sum_{q=\textrm{multiple of }m}^n \binom{n}{q} j^{n-q} m k^{lq/m}$

which is an integer. Call it $I$. Then

$\displaystyle \sum_{p=0}^{m-1} (j+e^{i2\pi p/m}k^{l/m})^n = (j+k^{l/m})^n + \sum_{p=1}^{m-1} (j+e^{i2\pi p/m}k^{l/m})^n = I$

or

$\displaystyle (j+k^{l/m})^n= I - \sum_{p=1}^{m-1} (j+e^{i2\pi p/m}k^{l/m})^n$

So for large $n$$(j+k^{l/m})^n$ approaches an integer if and only if the magnitudes of all the $m-1$ quantities $j+e^{i2\pi p/m}k^{l/m}$ have magnitude $<1$.

For instance: Let $j = 1$, $l = 1$. Then

$\displaystyle (1+k^{1/m})^n= I - \sum_{p=1}^{m-1} (1+e^{i2\pi p/m}k^{1/m})^n$

Now in the case $m = 2$,

$\displaystyle (1+k^{1/2})^n= I - (1+e^{i\pi}k^{1/2})^n = I - (1-k^{1/2})^n$

The magnitude of $1-k^{1/2}<1$ for $k<4$. In fact it’s zero for $k=1$, but then $1^{1/2}$ is an integer anyway; point is, $k = 2$ or $3$ works: $(1+\sqrt 2)^n$ and $(1+\sqrt 3)^n$ both approach integers (the former much more quickly than the latter).

How about $m = 3$? Then

$\displaystyle (1+k^{1/3})^n= I - (1+e^{i2\pi/3}k^{1/3})^n - (1+e^{i4\pi/3}k^{1/3})^n$

By symmetry the magnitudes of the two complex numbers are the same, so what we need is

$\displaystyle |1+e^{i2\pi/3}k^{1/3}|<1$

$\displaystyle 1 + k^{2/3} -k^{1/3}<1$

or

$\displaystyle k < 1$

So there are no integer values of $k$ that give convergence to an integer for $(1+k^{1/3})^n$. It seems evident the same is true for all $m>2$.