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


\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^{2\pi/3} + e {4\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


\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


\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.



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