Prove that if $p$ is an odd prime, then $$p\mid \lfloor(2+\sqrt5)^p\rfloor - 2^{p+1}$$

The solution posted by another user is as follows:

Let $𝑁=(2+\sqrt5)^p + (2-\sqrt5)^𝑝$. Note that $N$ is an integer. There are various ways to see this. One can, for example, expand using the binomial theorem, and observe that the terms involving odd powers of $\sqrt5$ cancel.

Because $(2−\sqrt5)^𝑝$ is a negative number close to $0$, it follows that $𝑁=⌊(2+\sqrt5)^𝑝⌋$.

In the two binomial expansions, all the binomial coefficients $\binom{p}{k}$ apart from the first and last are divisible by $p$. The first term in each expansion is $2^p$. We conclude that $N\equiv 2\cdot 2^p\pmod{p}$, and the result follows.

Could someone explain this proof in more detail please? I don't understand the following:

1) How the odd powers of $\sqrt5$ cancel,

2) How $N=\lfloor(2+\sqrt5)^p\rfloor$ (I understand that $(2-\sqrt5)^p$ is negative),

3) How $N\equiv 2\cdot 2^p\pmod{p}$, and the result follows.

How would one go about generalising to a result about $p^n$?


  • $\begingroup$ could you provide the link to the older question? $\endgroup$ – Dr. Mathva Feb 12 '19 at 14:00
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    $\begingroup$ $1)$: $\binom{p}{j}2^{(p-j)}(\sqrt{5}^j)+\binom{p}{j}2^{(p-j)}(-\sqrt{5})^j=0$ for odd $j$ $\endgroup$ – Peter Melech Feb 12 '19 at 14:05
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    $\begingroup$ math.stackexchange.com/questions/1666207/… $\endgroup$ – user437703 Feb 12 '19 at 14:09
  • $\begingroup$ More generally see PV (Pisot–Vijayaraghavan) numbers. $\endgroup$ – Bill Dubuque Feb 12 '19 at 15:17
  • $\begingroup$ I'd calculate by hand the numbers $(2 \pm \sqrt 5)^p$ for small odd primes up to maybe $p = 29$. A pattern would probably emerge. $\endgroup$ – Robert Soupe Feb 13 '19 at 3:31

1) We have
$(2+\sqrt 5)^p=2^p+\binom p12^{p-1}\sqrt 5+\binom p22^{p-2}\sqrt 5^2+\binom p23^{p-3}\sqrt 5^3+...+\binom pp\sqrt 5^p$, and $(2-\sqrt 5)^p=2^p-\binom p12^{p-1}\sqrt 5+\binom p22^{p-2}\sqrt 5^2-\binom p23^{p-3}\sqrt 5^3+...-\binom pp\sqrt 5^p$. When you add these together, all the odd terms are equal and opposite, so you get $N=2\times2^p+2\times\binom p22^{p-2}\sqrt 5^2+...+2\times\binom p{p-1}2\sqrt 5^{p-1}$.
Since this only has even powers of $\sqrt 5$, it is an integer.

2) As well as being an integer, $N$ is slightly less than $(2+\sqrt 5)^p$. In fact it is $-(2-\sqrt 5)^p=(\sqrt 5-2)^p$ less. Since $0<(\sqrt 5-2)^p<1$, $N$ is the integer between $(2+\sqrt 5)^p$ and $(2+\sqrt 5)^p-1$, i.e. it is $\lfloor(2+\sqrt 5)^p\rfloor$.

3) Going back to $N=2\times2^p+2\times\binom p22^{p-2}\sqrt 5^2+...+2\times\binom p{p-1}2\sqrt 5^{p-1}$, each term except the first has a factor of $\binom pr$ for some $0<r<p$. Writing this in factorial form, it is $\frac{p!}{r!(p-r)!}$. Since $p$ is prime, $p$ divides the top but not the bottom. This means every term except the first is a multiple of $p$, so $N\equiv 2\times2^p$ mod $p$.

  • $\begingroup$ Thanks for the reply. How would you go about generalising for $p^n$? My initial idea would be to use induction. $\endgroup$ – user437703 Feb 14 '19 at 17:27

For 1), the sum of odd powers $a^n + b^n$ can be factored into $(a+b)(a^{n-1}-a^{n-2}b+a^{n-3}b^2..........+b^{n-1})$ so where binomials $a$ and $b$ contain the same negative and positive irrational, these two factors will always be integers as positive and negative quantities of the irrational will be equal.

Example: $(2+\sqrt 5)^3 + (2 - \sqrt 5)^3 = (2+\sqrt 5+2 - \sqrt 5)((4 + 4\sqrt 5 + 5) - (4 - 5) + (4 - 4\sqrt 5 + 5)) = 4(19) = 76$


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