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[数论] 某abcd整除证明题劣化

$z^2|xy,z|x+y$
$\displaystyle x=\prod_i p_i^{x_i},y=\prod_i p_i^{y_i},z=\prod_i p_i^{z_i}$
$z^2|xy\Leftrightarrow \forall i(x_i+y_i\ge 2z_i)\Rightarrow
\forall i(x_i\ge z_i \lor y_i\ge z_i)$
$z|x+y,\forall i(x_i\ge z_i \lor y_i\ge z_i)\Rightarrow
\forall i(x_i\ge z_i,y_i\ge z_i)\Leftrightarrow z|x,z|y$
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本帖最后由 tommywong 于 2018-5-14 18:44 编辑

$\displaystyle \sigma_m(A)=\sum_{j_1<j_2<\dots<j_m\atop j_1,j_2,\dots,j_m\in A} x_{j_1}\dots x_{j_m}$

证明$d^m|\sigma_m(A),m=1,2,\dots,|A| \Leftrightarrow d|x_j,\forall j\in A$

n=1:

$d|\sigma_1(\{1\})=x_1 \Leftrightarrow d|x_1$

n=k+1:

$d^m|\sigma_m(\{1,2,\dots,k+1\}),m=1,2,\dots,k+1$

$d^{k+1}|\sigma_{k+1}(\{1,2,\dots,k+1\})=x_1 x_2\dots x_{k+1}$

$\displaystyle d=\prod_i p_i^{k_i}$

$\forall i,p^{k_i(k+1)}|x_1 x_2\dots x_{k+1}\Rightarrow \forall i,\exists r_i\in \{1,2,\dots,k+1\},s.t. p^{k_i}|x_{r_i}$

$\sigma_1(\{1,2,\dots,k+1\})=\sigma_1(\{1,2,\dots,k+1\}-\{r_i\})+x_{r_i}$

$d|\sigma_1(\{1,2,\dots,k+1\})\Rightarrow p^{k_i}|\sigma_1(\{1,2,\dots,k+1\})\Rightarrow p^{k_i}|\sigma_1(\{1,2,\dots,k+1\}-\{r_i\})=\sigma_1(\{1,2,\dots,k+1\})-x_{r_i}$

$\sigma_{m+1}(\{1,2,\dots,k+1\})=\sigma_{m+1}(\{1,2,\dots,k+1\}-\{r_i\})+x_{r_i}\sigma_m(\{1,2,\dots,k+1\}-\{r_i\})$

$d^{m+1}|\sigma_{m+1}(\{1,2,\dots,k+1\})\Rightarrow p^{k_i(m+1)}|\sigma_{m+1}(\{1,2,\dots,k+1\})$

$p^{k_i(m+1)}|\sigma_{m+1}(\{1,2,\dots,k+1\}),p^{k_i m}|\sigma_m(\{1,2,\dots,k+1\}-\{r_i\}),p^{k_i}|x_{r_i}\Rightarrow p^{k_i(m+1)}|\sigma_{m+1}(\{1,2,\dots,k+1\}-\{r_i\})=\sigma_{m+1}(\{1,2,\dots,k+1\})-x_{r_i}\sigma_m(\{1,2,\dots,k+1\}-\{r_i\})$

$\Rightarrow \forall i,p^{k_i m}|\sigma_m(\{1,2,\dots,k+1\}-\{r_i\}),m=1,2,\dots,k\Rightarrow \forall i,p^{k_i}|x_j,j\in \{1,2,\dots,k+1\}-\{r_i\}$

$\Rightarrow \forall i,p^{k_i}|x_j,j\in \{1,2,\dots,k+1\}\Rightarrow d|x_j,j\in \{1,2,\dots,k+1\}$

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回复 2# tommywong
写的什么哟,不忍卒读(实际是学的太渣)

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本帖最后由 tommywong 于 2018-4-11 18:48 编辑

看看这个实例

$z=6,x=8,y=9$满足$z^2|xy\Leftrightarrow 2^2 3^2|2^3 3^2$

这样不可能推出$z|x$或者$z|y$,实例也不成立

但可以推出x,y当中必有一个整除2、也必有一个整除3

这是因为若$x_i+y_i\ge 2z_i$,就不可能同时存在$x_i<z_i$、$y_i<z_i$的情况,必有一个大于或等于$z_i$

得到$p_i^{k_i}|x$或者$p_i^{k_i}|y$后,再加上$p_i^{k_i}|x+y$这个条件,其中一个成立将导致另外一个成立

于是就得出$p_i^{k_i}|x$、$p_i^{k_i}|y$同时成立

由于实例中的z并不整除x+y,所以就没有这个结论

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還可以怎樣玩呢?

$z^2|2xy,z|x+y$

$(p,2)=1$

$v_p(z^2)=2v_p(z)\le v_p(2xy)=v_p(x)+v_p(y)$

$v_p(x)\ge v_p(z) \lor v_p(y)\ge v_p(z)$

$v_p(x)\ge v_p(z) \land v_p(y)\ge v_p(z)$

$v_2(z^2)=2v_2(z)\le v_2(2xy)=1+v_2(x)+v_2(y)$

$v_2(x)\ge v_2(z) \lor v_2(y)\ge v_2(z)$

$v_2(x)\ge v_2(z) \land v_2(y)\ge v_2(z)$

$z|x,z|y$

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本帖最后由 tommywong 于 2018-5-14 21:13 编辑

$\displaystyle \sigma_m(A)=\sum_{j_1<j_2<\dots<j_m\atop j_1,j_2,\dots,j_m\in A} x_{j_1}\dots x_{j_m}$

若$\forall p,0\le v_p(c_m)\le m-1$

证明$d^m|c_m\sigma_m(A),m=1,2,\dots,|A| \Leftrightarrow d|x_j,\forall j\in A$

n=1:

$\forall p,v_p(c_1)=0\Rightarrow c_1=1$

$d|\sigma_1(\{1\})=x_1 \Leftrightarrow d|x_1$

n=k+1:

$d^m|c_m\sigma_m(\{1,2,\dots,k+1\}),m=1,2,\dots,k+1$

$d^{k+1}|c_{k+1}\sigma_{k+1}(\{1,2,\dots,k+1\})=c_{k+1}x_1 x_2\dots x_{k+1}$

$(k+1)v_p(d)\le v_p(c_{k+1})+v_p(x_1)+v_p(x_2)+\dots+v_p(x_{k+1})$

假设$\forall r\in \{1,2,\dots,k+1\},v_p(x_r)\le v_p(d)-1$

$(k+1)v_p(d)\le v_p(c_{k+1})+v_p(x_1)+v_p(x_2)+\dots+v_p(x_{k+1})\le (k+1)v_p(d)-1$ 矛盾

$\Rightarrow\forall p,\exists r_p\in \{1,2,\dots,k+1\},s.t. v_p(x_{r_p})\ge v_p(d)$

$\sigma_1(\{1,2,\dots,k+1\})=\sigma_1(\{1,2,\dots,k+1\}-\{r_p\})+x_{r_p}$

$d|c_1\sigma_1(\{1,2,\dots,k+1\})\Rightarrow
v_p(d)\le v_p(\sigma_1(\{1,2,\dots,k+1\}))$

$\Rightarrow
v_p(\sigma_1(\{1,2,\dots,k+1\}-\{r_i\}))=
v_p(\sigma_1(\{1,2,\dots,k+1\})-x_{r_p})\ge \inf\{v_p(d),v_p(d)\}=v_p(d)$

$\sigma_{m+1}(\{1,2,\dots,k+1\})=\sigma_{m+1}(\{1,2,\dots,k+1\}-\{r_p\})+x_{r_p}\sigma_m(\{1,2,\dots,k+1\}-\{r_p\})$

$d^{m+1}|c_{m+1}\sigma_{m+1}(\{1,2,\dots,k+1\})\Rightarrow (m+1)v_p(d)\le v_p(c_{m+1}\sigma_{m+1}(\{1,2,\dots,k+1\}))$

$\Rightarrow v_p(c_{m+1}\sigma_{m+1}(\{1,2,\dots,k+1\}-\{r_p\}))\ge \inf\{(m+1)v_p(d),v_p(x_{r_p})+mv_p(d)\}=(m+1)v_p(d)$

$\Rightarrow \forall p,mv_p(d)\le v_p(c_m\sigma_{m}(\{1,2,\dots,k+1\}-\{r_p\})),m=1,2,\dots,k$

$\Rightarrow \forall p,v_p(d)\le v_p(x_j),j\in\{1,2,\dots,k+1\}-\{r_p\}$

$\Rightarrow \forall p,v_p(d)\le v_p(x_j),j\in \{1,2,\dots,k+1\}$

$\Rightarrow d|x_j,j\in \{1,2,\dots,k+1\}$

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本帖最后由 tommywong 于 2018-5-14 21:09 编辑

$\begin{cases}
d|a+b+c\\
d^2|a^2+b^2+c^2\\
d^3|a^3+b^3+c^3
\end{cases}$

$d^2|(a+b+c)^2$

$d^2|a^2+b^2+c^2=(a+b+c)^2-2(ab+ac+bc)\Rightarrow d^2|2(ab+ac+bc)$

$d^3|a^3+b^3+c^3\Rightarrow d^3|2(a^3+b^3+c^3)=2(a+b+c)(a^2+b^2+c^2)-2(ab+ac+bc)(a+b+c)+6abc$

$d^3|(a+b+c)(a^2+b^2+c^2),d^3|2(ab+ac+bc)(a+b+c)\Rightarrow d^3|6abc$

$\displaystyle s_m=\sum_{i=1}^n x_i^m$

$d^m|s_m,m=1,2,\dots,n$

$d|s_1\Rightarrow d|\sigma_1$

$d^2|s_2\Rightarrow d^2|\sigma_1 s_1-2\sigma_2\Rightarrow d^2|2\sigma_2$

$\displaystyle s_{k+1}=(-1)^k (k+1)\sigma_{k+1}+\sum_{i=1}^k (-1)^{k+i}\sigma_{k+1-i} s_i$

$\displaystyle d^{k+1}|s_{k+1}\Rightarrow d^{k+1}|(-1)^k (k+1)\sigma_{k+1}+\sum_{i=1}^k(-1)^{k+i}\sigma_{k+1-i}s_i\Rightarrow d^{k+1}|(-1)^k (k+1)!\sigma_{k+1}+\sum_{i=1}^k (-1)^{k+i}k!\sigma_{k+1-i}s_i$

$d^i|s_i,d^{k+1-i}|(k+1-i)!\sigma_{k+1-i}\Rightarrow d^{k+1}|(k+1-i)!\sigma_{k+1-i} s_i\Rightarrow d^{k+1}|k!\sigma_{k+1-i}s_i\Rightarrow d^{k+1}|(k+1)!\sigma_{k+1}$

$\displaystyle v_p(m!)=\sum_{i=1}^\infty [\frac{m}{p^i}]\le m\sum_{i=1}^\infty \frac{1}{p^i}\le m\sum_{i=1}^\infty \frac{1}{2^i}<m\Rightarrow 0\le v_p(m!)\le m-1$

$d^m|s_m,m=1,2,\dots,n\Rightarrow d^m|m!\sigma_m,m=1,2,\dots,n\Rightarrow d|x_m,m=1,2,\dots,n$

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