11

I am looking for a proof on why $$\frac{1}{\sqrt 5}\left[\left(\frac{1+\sqrt 5}{2}\right)^n-\left(\frac{1-\sqrt 5}{2}\right)^n\right]$$ an integer.

I have seen many proofs on this, but they all refer to a properties of Fibonacci numbers, which shouldn't be necessary.

I am trying to see why it is true using purely elementary results such as the binomial formula. Clearly this reduces to $$\frac{1}{2^n\sqrt 5}\sum_{k=0}^n {n\choose k }\left(1-(-1)^k\right) 5^{k/2}$$ I am looking for a "divisibility" argument to see why this is an integer.

user593295
  • 562
  • 4
    It is not too hard to see it is rational; because $\frac{1 - \sqrt{5}}{2}$ is the conjugate of $\frac{1 + \sqrt{5}}{2}$, so the rational parts will cancel and $\left(\frac{1 + \sqrt5}{2}\right)^n - \left(\frac{1 - \sqrt5}{2}\right)^n$ will some rational number. Showing it is an integer takes a little more work. – Joseph Camacho Dec 10 '19 at 04:26
  • 2
    the rational part is obvious from the summation version. root(5) always cancels for odd k, and then terms are 0 for even k. How do you prove it is an integer however? – user593295 Dec 10 '19 at 04:28
  • 1
    Did you forget binomial coefficients in your summation? – J. W. Tanner Dec 10 '19 at 04:35
  • See https://math.stackexchange.com/questions/936479/proving-that-frac-phi4001-phi200-is-an-integer – lab bhattacharjee Dec 10 '19 at 04:58
  • 1
    I wouldn't jump to saying "which shouldn't be necessary" so quickly - the reason those show up is that the numbers $\frac{1\pm \sqrt{5}}2$ are the roots to $x^2=x+1$ and if you want to write $x^n=a_nx+b_n$ for rational $a_n,b_n$ you immediately find out that the algebraic equation leads to a recurrence relation - and this always happens for algebraic numbers. It only looks unnatural when you insist upon writing these $x$ in terms of some $y$ that satisfies $y^2=5$ (...which invokes the recurrence relation $A_{n+2}=5A_n$ to write powers of $y$ as rational combinations of $1$ and $y$) – Milo Brandt Dec 12 '19 at 06:07
  • Related : https://math.stackexchange.com/questions/1903099/why-is-2-sqrt3n2-sqrt3n-an-integer/1903132#1903132 – Watson Dec 22 '19 at 15:03

2 Answers2

9

One of the easiest proofs goes by induction.

  • Set $a= \frac{1+\sqrt{5}}{2}$ and $b= \frac{1-\sqrt{5}}{2}$

Then you have $$ab = -1 \mbox{ and } a+b = 1$$

Induction start $n= 1$: $\frac{1}{\sqrt 5}\left(a-b\right) = 1$ (Note, that for $n=0$ it is trivially true.)

Induction hypothesis: $\frac{1}{\sqrt 5}\left(a^k-b^k\right)$ is integer for $0\leq k \leq n$.

Induction step $n\to n+1$ ($n \geq 1$): \begin{eqnarray} \frac{1}{\sqrt 5}\left(a^{n+1}-b^{n+1}\right) & = & \frac{1}{\sqrt 5}\left((a+b-b)a^{n}-(b+a-a)b^{n}\right) \\ & = & \frac{1}{\sqrt 5}\left((a+b)(a^{n}-b^{n})-ba^n + ab^n \right) \\ & = & \underbrace{\frac{1}{\sqrt 5}\left((a+b)(a^{n}-b^{n})\right)}_{integer} + \frac{1}{\sqrt 5}\left(-\underbrace{ba}_{=-1}a^{n-1} + \underbrace{ab}_{=-1}b^{n-1} \right) \\ & = & \underbrace{\frac{1}{\sqrt 5}\left((a+b)(a^{n}-b^{n})\right)}_{integer} + \underbrace{\frac{1}{\sqrt 5}\left(a^{n-1} - b^{n-1} \right)}_{integer} \\ \end{eqnarray}

trancelocation
  • 32,243
  • Really really nice job (+1). Quick and easy – clathratus Dec 10 '19 at 05:14
  • Very nice, as for as elementary goes, this is it. I am curious to know if the binomial formula will do it as well? I feel as if it is even easier – user593295 Dec 10 '19 at 06:16
  • 2
    You're actually going from $n-1$ and $n$ to $n+1$, so you have to do two base cases, not just one, @trance. Much as I suggested in my answer. – Gerry Myerson Dec 12 '19 at 02:15
  • 1
    The binomial formula looks like a much harder way to try to solve this problem, user. – Gerry Myerson Dec 12 '19 at 02:16
  • @GerryMyerson : I considered the case $n= 0$ as trivially true and the induction step $n\to n+1$ started for me anyways at $n=1$. So, I skipped noting this. But of course you are right in mentioning that. So, I added this to my answer. – trancelocation Dec 12 '19 at 05:34
  • 1
    @user593295 That power of two in the denominator is kind of a problem when using the binomial formula. Gerry Myerson's answer gives a nice way of controlling it. Using $$ \left(\frac{1\pm\sqrt5}2\right)^3=2\pm\sqrt5$$ solves it when $n$ is a multiple of three, but leaves the other cases. – Jyrki Lahtonen Dec 12 '19 at 05:36
  • FWIW I decided to upvote this answer also. I do feel that it is, for many purposes, the same induction that you would have when solving this using Fibonacci recursion, but... shrug. I didn't see a better way of completing the trick with $3\mid n$ :-) – Jyrki Lahtonen Dec 12 '19 at 05:42
  • Sorry, I missed that you were taking as the induction hypothesis not just integer for $n$ but integer for everything up to $n$. – Gerry Myerson Dec 12 '19 at 06:03
8

Since $(1\pm\sqrt5)/2$ are both algebraic integers, so is $$\alpha=\left({1+\sqrt5\over2}\right)^n-\left({1-\sqrt5\over2}\right)^n$$ and since $\alpha$ is of the form $c\sqrt5$ with $c$ rational, $c$ must be an integer.

We use here this well-known fact: the algebraic integers of the form $(a+b\sqrt5)/2$ are precisely those where $a$ and $b$ are integers of the same parity.

If you insist on purely elementary methods, you can prove it for $n=0$ and $n=1$, and then use induction, after showing that the numbers for $n=k$ and for $n=k+1$ add up to the number for $n=k+2$.

Gerry Myerson
  • 179,216