数论 1

Quack 知道好多东西，把它们都做成 ppt。

inv_gcd 还可以用递推矩阵算。

void exgcd(int a, int b){
  r[0] = a, r[1] = b;
  int i = 2;
  for ( ; r[i - 1]; i++) {
    r[i] = r[i - 2] % r[i - 1];
    q[i - 1] = r[i - 2] / r[i - 1];
  }
  d = r[i -= 2];
  mat res(1, 0, 0, 1);
  for (int j = i - 1; j; --j)
    res.mul(mat(0, 1, 1, -q[j]));
  s = res.get(1, 0); t = res.get(1, 1);
}

还有个 trick 叫 binary gcd。

int gcd(int a, int b){
  if (a == b) return a;
  if ((a & 1) && (b & 1)) return gcd(b, abs(a - b));
  else if ((a & 1) && (!(b & 1))) return gcd(a, b >> 1);
  else if ((!(a & 1)) && (b & 1)) return gcd(a >> 1, b);
  else return gcd(a >> 1, b >> 1) << 1;
}

复杂度和正确性平凡吧，在写高精度 gcd 的时候可以考虑。接下来看一个典中典问题：

证明：\((a^n-1, a^m-1) = a^{(n, m)}-1\)。

P.f.：不妨设 \(n \leqslant m\)，首先 \((a^n-1, a^m-1) = (a^n(a^{m-n}-1), a^n-1) = (a^{m-n}-1, a^n-1)\)，一直迭代可以得到 \((a^n-1, a^m-1) = (a^{(n, m)}-1, a^{(n, m)}-1) = a^{(n, m)}-1\)。

然后这个有个 more general 的版本，就是对于任意整数 \(a, b\)，满足 \((a, b) = 1\) 则有 \((a^n-b^n, a^m-b^m) = a^{(n, m)}-b^{(n, m)}\)，证明是类似的。

但是如果说我们还有一个 more more general 的版本呢？

Theorem：若 \(f_n\) 是一个整数序列，并满足 \(f_0 = 0\)，\(f_{n} \equiv f_{n-m} \pmod {f_m},n > m\)，有 \((f_n, f_m) = f_{(n, m)}\)。

P.f.：考虑 induction，当 \(n = m\)，\(n = 0\)，\(m = 0\) 时，上述命题为一个，一个一个一个，一个一个一个一个一个一个一个

什么栈什么溢什么出上还有一些 q-analog 的高论，不过我连上面的证明都还没看懂。引一下原 writer：

Mimic in expts a subtractive Euclidean algorithm \(\rm\,(n,m) = (\color{#0a0}{n\!-\!m},m)\)

\[\begin{align} \rm{e.g.}\ \ &\rm (f_5,f_2) = (f_3,f_2) = (f_1,f_2) = (f_1,f_1) = (f_1,\color{darkorange}{f_0})= f_1 = f_{\:\!(5,\,2)}\\[.3em] {\rm like}\ \ \ &(5,\ 2)\, =\:\! (3,\ 2)\, =\:\! (1,\ 2)\:\! =\:\! (1,\ 1)\:\! =\:\! (1,\ \color{darkorange}0)\:\! = 1,\ \ {\rm since}\end{align}\qquad\]

$\rm\ f_{,n}: :=\ a^{n!-!1 = a} : \color{#c00}{(a^m!-!1)} + \color{#0a0}{a^{n-m}!-!1},,\ $ hence \(\rm\:\ {f_{\,n}\! = \color{#0a0}{f_{\,n-m}}\! + k\ \color{#c00}{f_{\,m}}},\,\ k\in\mathbb Z,\:\) thus

Theorem $: $ If $\rm\ f_{, n}: $ is an integer sequence with $\rm\ \color{darkorange}{f_{0} =, 0},: $ $\rm :{ f_{,n}!\equiv \color{#0a0}{f_{,n-m}}\ (mod\ \color{#c00}{f_{,m})}}\ $ for all $\rm: n > m,\ $ then $\rm: (f_{,n},f_{,m})\ =\ f_{,(n,:m)}, : $ where $\rm\ (i,:j)\ $ denotes \(\rm\ gcd(i,\:j).\:\)

Proof $\ $ By induction on \(\rm\:n + m\:\). The theorem is trivially true if $\rm\ n = m\ $ or $\rm\ n = \color{darkorange}0\ $ or \(\rm\: m = \color{darkorange}0.\:\)
So we may assume \(\rm\:n > m > 0\:\).$\ $ Note $\rm\ (f_{,n},f_{,m}) = (\color{#0a0}{f_{,n-m}},\color{#c00}{f_{,m}})\ $ follows by \(\rm\color{#90f}{Euclid}\) & hypothesis.
Since $\rm\ (n-m)+m \ <\ n+m,\ $ induction yields \(\rm\, \ (f_{\,n-m},f_{\,m})\, =\, f_{\,(n-m,\:m)} =\, f_{\,(n,\:m)}.\)

\(\rm\color{#90f}{Euclid}\!:\ A\equiv a\pmod{\! m}\,\Rightarrow\ (A,m) = (a,m)\,\) is the reduction (descent) step used both above and in the Euclidean algorithm $\rm: (A,m) = (A\bmod m,,m),, $ the special case \(\,\rm f_{\:\!n} = n\,\) above.

This is a prototypical strong divisibility sequence. Same for Fibonacci numbers.

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Alternatively it has a natural proof via the Order Theorem \(\ a^k\equiv 1\iff {\rm ord}(a)\mid k,\,\) viz.

\[\begin{eqnarray}\ {\rm mod}\:\ d\!:\ a^M\!\equiv 1\equiv a^N&\!\iff\!& {\rm ord}(a)\ |\ M,N\!\color{#c00}\iff\! {\rm ord}(a)\ |\ (M,N)\!\iff\! \color{#0a0}{a^{(M,N)}\!\equiv 1}\\[.2em] {\rm i.e.}\ \ \ d\ |\ a^M\!-\!1,\:a^N\!-\!1\! &\!\iff\!\!&\ d\ |\ \color{#0a0}{a^{(M,N)}\!-\!1},\qquad\,\ {\rm where} \quad\! (M,N)\, :=\, \gcd(M,N) \end{eqnarray}\ \ \ \ \ \]

Thus, by above $, a^M!-!1,:aN!-!1\ $ and $, a^{(M,N)}!-!1\ $ have the same set \(\,S\,\) of common divisors $,d,, $ therefore they have the same greatest common divisor \(\ (= \max\ S).\)

Note We used the GCD universal property $\ a\mid b,c \color{#c00}\iff a\mid (b,c)\ $ [which is the definition of a gcd in more general rings]. Compare that with $\ a<b,c !\iff! a< \min(b,c),, $ and, analogously, $,\ a\subset b,c\iff a\subset b\cap c.\ $ Such universal "iff" characterizations enable quick and easy simultaneous proof of both directions.

The conceptual structure that lies at the heart of this simple proof is the ubiquitous order ideal. $\ $ See this answer for more on this and the more familiar additive form of a denominator ideal.