Proof Minimal Polynomial of $\left (\sqrt{2} + \sqrt{3} \right ) ^{666}$ in $\mathbb{Z}$

Question

Wolfram says it is: $$ p:=X^2 - \\34010363120518329895641913244818972821357722422049225139833901083899949664\\3109676116279151909642219515106579657522329002315790814575298372173755500\\4450204396353513751397454807621451130987201489527449320182179843543647401\\3457441914149811405514490266436995563308519019060191654063799652512313885\\722205521300795188376507318743522032330 \cdot X + 1$$

Obviously the degree cannot be any lower: Otherwise $z:=\left (\sqrt{2} + \sqrt{3} \right )^{666} \in \mathbb {Z}$ .

However I have trouble proving that $p(z)=0$.

Here is what I tried so far:

There are $a_n \in \left \{ 0 \leq a <10 \space :\space a \in \mathbb{N} \right \}$ and $k\in\mathbb{N}$ so that $$z=\sum_{n=-k}^{\infty}a_n \cdot 10^{-n} $$

Define:$$z_1:=\sum_{n=-k}^{0}a_n \cdot 10^{-n}$$ $$z_2:=\sum_{n=1}^{\infty}a_n \cdot 10^{-n}$$

Coincidentally(?) the second coeffecient of p is exactly $-\left ( z_1+1\right )$: $p(z)=p(z_1+z_2)=\left ( z_1+z_2 \right )^2-\left ( z_1+1 \right ) \left ( z_1+z_2 \right )+1=\left ( z_1+z_2 \right ) \left ( z_2 - 1 \right ) + 1= z\left (z_2-1 \right ) +1 =? $

Any ideas, why $\left (1-z_1 \right )$ could be the multiplicative inverse of z?

Am I on the right track at all?

Answer 1

Define $a_n = (\sqrt{2} + \sqrt{3})^{2n}$ for $n \in \mathbb{N}$. Because\begin{align*} a_n + \frac{1}{a_n} &= (\sqrt{3} + \sqrt{2})^{2n} + (\sqrt{3} - \sqrt{2})^{2n}\\ &= (5 + 2\sqrt{6})^n + (5 - 2\sqrt{6})^n\\ &= \sum_{k = 0}^n \binom{n}{k} 5^{n - k} (2\sqrt{6})^k + \sum_{k = 0}^n \binom{n}{k} 5^{n - k} (-2\sqrt{6})^k\\ &= 2 \sum_{\substack{0 \leqslant k \leqslant n\\k \ \text{even}}}\binom{n}{k} 5^{n - k} (2\sqrt{6})^k = 2 \sum_{0 \leqslant 2k \leqslant n} \binom{n}{2k} 5^{n - 2k} 24^k \in \mathbb{Z}, \end{align*} then$$ a_n^2 - 2 \sum_{0 \leqslant 2k \leqslant n} \binom{n}{2k} 5^{n - 2k} 24^k \cdot a_n + 1 = 0. $$ Since $a_n \notin \mathbb{Q}$, the minimal polynomial of $a_n$ in $\mathbb{Z}[x]$ is$$ X^2 - 2 \sum_{0 \leqslant 2k \leqslant n} \binom{n}{2k} 5^{n - 2k} 24^k \cdot X + 1. $$

Now, the question is the case where $n = 333$. Note that, by the definition of $z_1$, $z_1 = [a_n]$. Since $a_n > 1$ and $\displaystyle a_n + \frac{1}{a_n} \in \mathbb{Z}$, then$$ z_1 = a_n + \frac{1}{a_n} - 1 \Longrightarrow -\left(a_n + \frac{1}{a_n}\right) = -(z_1 + 1). $$

Answer 2

Note that $(\sqrt{2}+\sqrt{3})^2=5+2\sqrt{6}.$ So we can write $(5+2\sqrt{6})^{333}$ as $a+b\sqrt{6}$ and the other root of the minimal polynomial will be $a-b\sqrt{6}$. You'll get that the product of the roots must be $1$ and the sum of the two roots must be $2a$. So the minimal polynomial will be: $x^2-2ax+1.$ Working out what $a$ will be is a bit tedious. You can actually write:

$$2a=\left\lceil \left(5+2\sqrt{6}\right)^{333}\right\rceil$$

Or you can use the method of repeated squaring to compute $(5+2\sqrt 6)^{333}$ somewhat efficiently. (Certainly more efficiently than the summation formula below, which is fairly expensive.)

Here's a python script which computes the value of $2a:$

def prod(v1,v2):
   # Compute (v1[0]+v1[1]sqrt(6))(v2[0]+v2[1]sqrt(6))
   return (v1[0]*v2[0]+6*v1[1]*v2[1],v1[0]*v2[1]+v1[1]*v2[0])

value = (1,0)
n = 333
a_2_to_k = (5,2)

while (n>0):
   if n%2==1:
       value = prod(value,a_2_to_k)
   n = n/2
   a_2_to_k = prod(a_2_to_k,a_2_to_k)

print 2*value[0]

with output:

$$ 34010363120518329895641913244818972821357722422049\\ 22513983390108389994966431096761162791519096422195\\ 15106579657522329002315790814575298372173755500445\\ 02043963535137513974548076214511309872014895274493\\ 20182179843543647401345744191414981140551449026643\\ 69955633085190190601916540637996525123138857222055\\ 21300795188376507318743522032330$$