# Why do binomial expansions involving surds get closer to integers as they get larger? [duplicate]

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?

• What happens if you add $(2-\sqrt{3})^n$ to it? Sep 9, 2020 at 9:32
• 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. Sep 9, 2020 at 10:57
• How large is $(2-\sqrt{3})^n$? Sep 9, 2020 at 11:12
• Ohhhhh. (< 1) That's genius. Many thanks! Sep 9, 2020 at 11:22
• 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). Sep 10, 2020 at 23:24

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.