不定方程 $x^4+75y^4=z^2$ 中若设 $y\neq 0$, 则无正整数解.

若设 $(x,y)=1$, 则可推出 $(x,z)=1$.

斐波那契数列 Fibonacci(n), 或简记为 $F(n)$,

定义为 $F(n)=F(n-1)+F(n-2)$. $F(0)=0$, $F(1)=1$. 且 $F(-n)=(-1)^{n-1}F(n)$. 这里 $n$ 为非负整数.

Fibonacci 数列的前几个值为:
 \[0,1,1,2,3,5,8,13,21,34,55,\cdots\]

在 Calculator 中可以使用下面的函数生成

 Fibonacci(10,yes)

通项 F_n=F(n) 的公式为:

\[F_n = \frac{1}{\sqrt{5}}\Bigl(\frac{1+\sqrt{5}}{2}\Bigr)^n-\frac{1}{\sqrt{5}}\Bigl(\frac{1-\sqrt{5}}{2}\Bigr)^n\]

令

\[
m=\begin{cases}
n, & n \text{是奇数}\\
n-1, & n \text{是偶数}\\
\end{cases}
\]

则可推出

\[F_n = \frac{1}{2^{n-1}}\cdot(C_n^1+C_n^3\cdot 5+C_n^5\cdot 5^2+\cdots+C_n^m\cdot 5^{\frac{m-1}{2}})\]

Fibonacci 数和二项式之间还有如下关系 ( [1] P.121 题 28) :

\[
F_{n+1}=C_n^0+C_{n-1}^1+\cdots+C_{n-k}^{k},
\]

其中 $k=[\frac{n}{2}]$.

这也是 Fibonacci 数与 Pascal 三角形之间的关系.

References:

[1] 殷剑宏 编著 《组合数学》

素数 $p$ 称为是正则素数(regular prime), 如果它不能整除 $h_p$. 此处 $h_p$ 是指 cyclotomic field $\mathbb{Q}(\zeta_p)$ 中理想的等价类个数.

References:

Unsolved Problems in Number Theory, D2 The Fermat problem.

关于 $10^n+1$ 的因子分解

列出这样一些特殊的素数, $p=d_1d_2\ldots d_n$ 是素数, 它的反转(reverse) $q=\mathrm{reverse}(p)=d_n d_{n-1}\ldots d_2 d_1$ 也是素数.

比如 91121 和 12119

使用 Calculator 进行验证.

>> isprime(91121)
in> isprime(91121)
This is function isprime
sqrtnum=301.862552
--------------------
we choose sqrtnum : 303
 ... since we use the traditional algorithm, ...
 ... please wait a minute ...
 ... ::: ... ::: ...
out> 91121 is a prime

------------------------

>> reverse(91121)
in> reverse(91121)
out> 12119

>> isprime(12119)
in> isprime(12119)
This is function isprime
sqrtnum=110.086330
--------------------
we choose sqrtnum : 111
 ... since we use the traditional algorithm, ...
 ... please wait a minute ...
 ... ::: ... ::: ...
out> 12119 is a prime

(以下使用 Calculator 计算)

>> setprecision(200)
in> setprecision(200)
Now the precision is: 200

------------------------

>> continued_fraction(1,2,...,100)
in> continued_fraction(1,2,...,100)
out> 210628326606030925509504995396115751215082237173770949560136689651065020446811215535403721603460152269102594085067245222090841983806373539433329769060736986251|146971108555054591263734471078160507352050045572198741030186489794108431298591736983812285328810821610957278981960240168856854819813194627616076643581220741550

Expression:
1+1/(2+1/(3+1/(4+1/(5+1/(6+1/(7+1/(8+1/(9+1/(10+1/(11+1/(12+1/(13+1/(14+1/(15+1/(16+1/(17+1/(18+1/(19+1/(20+1/(21+1/(22+1/(23+1/(24+1/(25+1/(26+1/(27+1/(28+1/(29+1/(30+1/(31+1/(32+1/(33+1/(34+1/(35+1/(36+1/(37+1/(38+1/(39+1/(40+1/(41+1/(42+1/(43+1/(44+1/(45+1/(46+1/(47+1/(48+1/(49+1/(50+1/(51+1/(52+1/(53+1/(54+1/(55+1/(56+1/(57+1/(58+1/(59+1/(60+1/(61+1/(62+1/(63+1/(64+1/(65+1/(66+1/(67+1/(68+1/(69+1/(70+1/(71+1/(72+1/(73+1/(74+1/(75+1/(76+1/(77+1/(78+1/(79+1/(80+1/(81+1/(82+1/(83+1/(84+1/(85+1/(86+1/(87+1/(88+1/(89+1/(90+1/(91+1/(92+1/(93+1/(94+1/(95+1/(96+1/(97+1/(98+1/(99+1/(100)))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))

TeX Code:
1+\frac{1}{2+\frac{1}{3+\frac{1}{4+\frac{1}{5+\frac{1}{6+\frac{1}{7+\frac{1}{8+\frac{1}{9+\frac{1}{10+\frac{1}{11+\frac{1}{12+\frac{1}{13+\frac{1}{14+\frac{1}{15+\frac{1}{16+\frac{1}{17+\frac{1}{18+\frac{1}{19+\frac{1}{20+\frac{1}{21+\frac{1}{22+\frac{1}{23+\frac{1}{24+\frac{1}{25+\frac{1}{26+\frac{1}{27+\frac{1}{28+\frac{1}{29+\frac{1}{30+\frac{1}{31+\frac{1}{32+\frac{1}{33+\frac{1}{34+\frac{1}{35+\frac{1}{36+\frac{1}{37+\frac{1}{38+\frac{1}{39+\frac{1}{40+\frac{1}{41+\frac{1}{42+\frac{1}{43+\frac{1}{44+\frac{1}{45+\frac{1}{46+\frac{1}{47+\frac{1}{48+\frac{1}{49+\frac{1}{50+\frac{1}{51+\frac{1}{52+\frac{1}{53+\frac{1}{54+\frac{1}{55+\frac{1}{56+\frac{1}{57+\frac{1}{58+\frac{1}{59+\frac{1}{60+\frac{1}{61+\frac{1}{62+\frac{1}{63+\frac{1}{64+\frac{1}{65+\frac{1}{66+\frac{1}{67+\frac{1}{68+\frac{1}{69+\frac{1}{70+\frac{1}{71+\frac{1}{72+\frac{1}{73+\frac{1}{74+\frac{1}{75+\frac{1}{76+\frac{1}{77+\frac{1}{78+\frac{1}{79+\frac{1}{80+\frac{1}{81+\frac{1}{82+\frac{1}{83+\frac{1}{84+\frac{1}{85+\frac{1}{86+\frac{1}{87+\frac{1}{88+\frac{1}{89+\frac{1}{90+\frac{1}{91+\frac{1}{92+\frac{1}{93+\frac{1}{94+\frac{1}{95+\frac{1}{96+\frac{1}{97+\frac{1}{98+\frac{1}{99+\frac{1}{100}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}

------------------------

>> 210628326606030925509504995396115751215082237173770949560136689651065020446811215535403721603460152269102594085067245222090841983806373539433329769060736986251/146971108555054591263734471078160507352050045572198741030186489794108431298591736983812285328810821610957278981960240168856854819813194627616076643581220741550
in> 210628326606030925509504995396115751215082237173770949560136689651065020446811215535403721603460152269102594085067245222090841983806373539433329769060736986251/146971108555054591263734471078160507352050045572198741030186489794108431298591736983812285328810821610957278981960240168856854819813194627616076643581220741550

out> 1.43312742672231175831718345577599182043151276790598052343442863639430918325417290013650372643578611465950013404308853642953017708273894637360407321952533635247368315637151340965862626563444808561719787

------------------------

其他参考资料

https://math.stackexchange.com/questions/69519/closed-form-for-a-pair-of-continued-fractions

设 $g|ab$, $g|cd$ 及 $g|(ac+bd)$, 证明: $g|ac$ 且 $g|bd$.

证明: $x^2+2y^2=203$ 无整数解.

References:

潘承洞, 潘承彪, 《数论》   P. 

设 $N$ 不是平方数, 证明 $\sqrt{N}$ 是无理数.

事实上, 只要 $N$ 不是某个整数 $n$ 的 $m$ 次幂, 则 $\sqrt[m]{N}$ 都是无理数.

现在, 若 $\sqrt{N}$ 是无理数, 问是否有无理性的几何证明?

哈代数论中给出了 $\sqrt{5}$ 无理性的几何证明.

\[
\begin{split}
&(1+2x+2x^4+\cdots)^6\\
=&1+16\biggl(\frac{1^2 x}{1+x^2}+\frac{2^2 x^2}{1+x^4}+\frac{3^2 x^3}{1+x^6}+\cdots\biggr)-4\biggl(\frac{1^2 x}{1-x}-\frac{3^2 x^3}{1-x^3}+\frac{5^2 x^5}{1-x^5}-\cdots\biggr)
\end{split}
\]

\[
(1+2x+2x^4+\cdots)^8=1+16\biggl(\frac{1^3 x}{1+x}+\frac{2^3 x^2}{1-x^2}+\frac{3^3 x^3}{1+x^3}+\cdots\biggr)
\]

References:

哈代数论（第6版）20.13 用多个平方和表示数

G. H. Hardy, E. M. Wright, An Introduction to the Theory of Numbers.