Suppose I have a binomial expansion of the form: $$ (2+\sqrt{3})^n $$ Why is it that as $n$ approaches $\infty$ that the value of the expansion becomes closer and closer to being an integer?

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    $\begingroup$ What happens if you add $(2-\sqrt{3})^n$ to it? $\endgroup$ Sep 9, 2020 at 9:32
  • $\begingroup$ All the surd terms cancel out, and it is left as an integer. I can't see how to relate this to the above, however. $\endgroup$
    – Teymour
    Sep 9, 2020 at 10:57
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    $\begingroup$ How large is $(2-\sqrt{3})^n$? $\endgroup$ Sep 9, 2020 at 11:12
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    $\begingroup$ Ohhhhh. (< 1) That's genius. Many thanks! $\endgroup$
    – Teymour
    Sep 9, 2020 at 11:22
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    $\begingroup$ Dupe: Does this answer your question? Why is $(\sqrt{2}+\sqrt{3})^{2008}$ so close to an integer?. See also the questions linked there, and their links... (it's a FAQ). $\endgroup$ Sep 10, 2020 at 23:24

2 Answers 2


If $\alpha_1$ is an algebraic integer (which $2 + \sqrt{3}$ is) then it's the root of a monic irreducible polynomial $f(x) = x^d + \dots$ with integer coefficients, which here is

$$f(x) = (x - (2 + \sqrt{3}))(x - (2 - \sqrt{3})) = (x - 2)^2 - 3 = x^2 - 4x + 1.$$

This polynomial has some other roots $\alpha_2, \dots \alpha_d$, the conjugates of $\alpha_1$, and then you can show in various ways that:

Claim: The sequence $$p_n = \sum_{k=1}^d \alpha_k^n$$ of power sums is always an integer.

Here this sequence is $(2 + \sqrt{3})^n + (2 - \sqrt{3})^n$ as Jaap says in the comments. This is easiest to understand in the quadratic case $d = 2$ but it holds more generally.

If it further happens that the other roots $\alpha_2, \dots \alpha_d$ all have absolute value less than $1$, then their contributions to the power sum above decay exponentially as $n \to \infty$, and then for $n$ large enough that the sum of these contributions is less than $\frac{1}{2}$ (which happens quite quickly), $p_n$ will be the closest integer to $\alpha_1^n$. The real algebraic integers with this property are called Pisot-Vijayaraghavan numbers and they're somewhat rare but they do exist. The most famous one is probably the golden ratio $\phi = \frac{1 + \sqrt{5}}{2}$, whose conjugate is the "other" golden ratio $\varphi = \frac{1 - \sqrt{5}}{2}$. The sequence of power sums

$$L_n = \phi^n + \varphi^n$$

is the Lucas numbers, a close cousin of the more famous Fibonacci numbers, and $|\varphi^n| < \frac{1}{2}$ for $n \ge 2$ so we get that

Claim: For $n \ge 2$, $L_n$ is the closest integer to $\phi^n$.

There's an analogous formula for the Fibonacci numbers which goes

$$F_n = \frac{\phi^n - \varphi^n}{\phi - \varphi}$$

and similarly it implies

Claim: For $n \ge 1$, $F_n$ is the closest integer to $\left[ \frac{\phi^n}{\sqrt{5}} \right]$.

$2 + \sqrt{3}$ has this same kind of relationship to the sequence

$$p_n = (2 + \sqrt{3})^n + (2 - \sqrt{3})^n$$

which can (this is one of the ways to prove it always consists of integers) equivalently be defined as the sequence satisfying $p_0 = 2, p_1 = 4$ and the recurrence relation

$$p_{n+2} = 4 p_{n+1} - p_n.$$

This sequence begins $2, 4, 14, 52, \dots$ and I don't think it has a name but it's A003500 in the OEIS.


I was going to post this as another question, but I know the answer now and I think it's better to post it here. My question was going to be:

Can we use this fact: "Binomial expansions of some surds get closer to an integer as $n \to \infty$" to get arbitrarily good rational approximations of those surds?

The $2$ in the question of this thread may muddy the waters, so let's use another example: $(\sqrt13 + 3)^n$.

$(\sqrt13 + 3)^n + (\sqrt13 - 3)^n = 2[ \ (\sqrt13)^n + \binom{n}{2}(\sqrt13)^{n-2}(-3)^2 + ... + \binom{n}{n-2}(\sqrt13)^{2}(-3)^{n-2} + (-3)^n\ ] \implies$

$(\sqrt13 + 3)^n - 2[ \ (\sqrt13)^n + \binom{n}{2}(\sqrt13)^{n-2}(-3)^2 + ... + \binom{n}{n-2}(\sqrt13)^{2}(-3)^{n-2} + (-3)^n\ ] = (\sqrt13 - 3)^n$

$\to 0$ as $n \to \infty$, which was the little calculation Jaap Scherphuis indicated towards in his comment.

I guess the answer to my question is yes: consider when $n$ is a large and even number, pretend the right hand side is $0$ (which is appropriate) and re-arrange.

(Maybe it also works for odd $n$, but you don't need to consider that now that you can see it works for even $n$ ).

I'm not sure of the convergence rate of this approximation to rationals compared to some other methods, but perhaps that's an investigation for another day.


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