(1+√2)^n

hbghlyj

$x_n$ oeis.org/A001333
$y_n$ oeis.org/A000129
《简明数论》P282
6. 设 $x_n+y_n \sqrt{2}=(1+\sqrt{2})^n$. 证明：
(i) $y_{n+1}=x_n+y_n, \quad x_{n+1}=y_{n+1}+y_n, \quad n \geqslant 1$.
(ii) $y_{2 n+1}=y_{n+1}^2+y_n^2, \quad n \geqslant 1$.
(iii) $y_{2 n+1}^2$ 是两个相邻自然数的平方和, 求出这两个自然数.
(iv) 设 $x_0=1, y_0=0$. 当 $n \geqslant m$ 时,
\begin{align*}
x_n x_m-2 y_n y_m&=(-1)^m x_{n-m}\\
x_n y_m-y_n x_m&=(-1)^{m+1} y_{n-m}
\end{align*}
(v)
$$
\begin{aligned}
& x_{2 n+1}=x_{n+1} x_n+2 y_{n+1} y_n=2 x_{n+1} x_n+(-1)^{n+1}, \\
& y_{2 n+1}=x_{n+1} y_n+y_{n+1} x_n .
\end{aligned}
$$
(vi) $2 \mid y_{2 n}, 2 \nmid y_{2 n+1}$.
(vii) 当 $n>1$ 时, $x_n$ 不是完全平方数.

hbghlyj

(i) $x_{n+1}+y_{n+1}\sqrt{2}=(x_n+y_n \sqrt{2})(1+\sqrt{2})\implies \begin{aligned}y_{n+1}&=x_n+y_n\\x_{n+1}&=2y_n+x_n=y_{n+1}+y_n\end{aligned}$
(ii) 记$a=1+\sqrt2,b=1-\sqrt2,y_n=(a^n-b^n)/2\sqrt2$\begin{align*}
y_{n+1}^{2} + y_{n}^{2} &= \frac{1}{8} \left[ (a^{n+1} - b^{n+1})^{2} + (a^{n} - b^{n})^{2} \right] \\
&= \frac{1}{8} \left[ 2(2+\sqrt{2}) \, a^{2n} + 2 (2 - \sqrt{2}) \, b^{2n} \right] \\
&= \frac{a^{2n+1} - b^{2n+1}}{2 \sqrt{2}} \\
&= y_{2n+1}
\end{align*}
(iii) $y_{2n+1}^2=\left(x_{2 n+1}-1\over2\right)^2+\left(x_{2 n+1}+1\over2\right)^2$
等价于$2y_{2n+1}^2=x_{2 n+1}^2+1$\begin{align*}
2y_{2n+1}^2-x_{2 n+1}^2&=2\frac{(a^{2 n+1}-b^{2 n+1})^2}8-\frac{(a^{2 n+1}+b^{2 n+1})^2}4\\
&=\frac{(a^{2 n+1}-b^{2 n+1})^2-(a^{2 n+1}+b^{2 n+1})^2}4\\
&=-a^{2 n+1}b^{2 n+1}=-(-1)^{2 n+1}=1
\end{align*}
(iv)\begin{align*}
x_n x_m-2 y_n y_m&=\frac14[(a^n+b^n)(a^m+b^m)-(a^n-b^n)(a^m-b^m)]\\&=\frac12(a^nb^m+a^mb^n)\\&=(ab)^m\frac{a^{n-m}+b^{n-m}}2\\
&=(-1)^m x_{n-m}\\[1ex]
x_n y_m-y_n x_m&=\frac1{4\sqrt2}[(a^n+b^n)(a^m-b^m)-(a^n-b^n)(a^m+b^m)]
\\&=\frac1{2\sqrt2}(a^mb^n-a^nb^m)
\\&=(ab)^m\frac{b^{n-m}-a^{n-m}}{2\sqrt2}
\\&=-(-1)^m y_{n-m}
\end{align*}(v)\begin{align*}
2 x_{n+1} x_{n} &= \frac{1}{2} \, (a^{n+1} + b^{n+1})(a^{n} + b^{n}) \\
&= \frac{1}{2} ( a^{2n+1} + b^{2n+1}) + \frac{1}{2} (ab)^{n} (a+b) \\
&= x_{2n+1} + (-1)^{n}
\end{align*}

hbghlyj

$x_n^2-2y_n^2=-1\iff (y_n^2)^4-(\sqrt{x_n})^4=(y_n^4-1)^2$.
And this equation is the form of $x^4-y^4=z^2$ and it's famous that it only has solutions (1,1)

Proof

Suppose the equation has a solution in positive integers, and choose a solution that minimizes $x^2+y^2$. Note that $x,y$, and $z$ are pairwise coprime, since otherwise we could divide out by their common divisor to get a smaller solution. Thus
\[z^2 + (y^2)^2 = (x^2)^2\]
so that $z, y^2, x^2$ form a pythagorean triple. There are thus positive integers $p,q$ of opposite parity (and coprime since $x, y$, and $z$ are) such that $x^2 = p^2+q^2$ and either $y^2 = p^2-q^2$ or $y^2 = 2pq$.

Factoring the original equation, we get
\[(x^2-y^2)(x^2+y^2)=z^2\]
If $y^2 = p^2-q^2$, then $(xy)^2 = p^4-q^4$, and clearly $p^2+q^2 = x^2 < x^2+y^2$, so we have found a solution smaller than the assumed minimal solution.

Assume therefore that $y^2 = 2pq$. Now, $x^2 = p^2+q^2$; we may assume by relabeling if necessary that $q$ is even and $p$ odd. Then $p, q, x$ are pairwise coprime and form a pythagorean triple; thus there are $P>Q>0$ of opposite parity and coprime such that
\[
  q = 2PQ,\quad p=P^2-Q^2,\quad x=P^2+Q^2
\]
Then
\[
  PQ(P^2-Q^2) = \frac{1}{2}pq = \frac{y^2}{4}
\]
is a square; it follows that $P, Q$, and $P^2-Q^2$ are all (nonzero) squares since they are pairwise coprime. Write
\[
  P = R^2,\quad Q=S^2,\quad P^2-Q^2=T^2
\]
for positive integers $R,S,T$. Then $T^2=R^4-S^4$, and
\[
  R^2+S^2 = P+Q < (P+Q)(PQ)(P-Q) = \frac{1}{2}pq = \frac{y^2}{4}\leq y^2 < x^2+y^2
\]
We have thus found a smaller solution in positive integers, contradicting the hypothesis.