blocks|key|2425551|text|我试着用更多的细节来解释它。假设我们有下面的模方程。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2425552|2425553|a+\equiv+b+\mod(c)|atomic|offset|length|style|CODE|mathjax|teX|a+\equiv+b+\mod(c)|2425554|2425555|对于一个整数k，|2425556|2425557|a=b%2Bk+\cdot+c|a=b%2Bk+\cdot+c|2425558|2425559|例如，在\mod(26)中，对28的提醒等于2，对于包括\{2,28,44,80,...\}在内的所有2%2Bk+\cdot+26也是相同的提醒。在这一点上，我们需要弄清楚什么是加法和乘法。所以我声称，|2425560|2425561|a_1+\mod(c)%2B+a_2+\mod(c)\equiv+(a_1%2Ba_2)+\mod(c)|a_1+\mod(c)%2B+a_2+\mod(c)\equiv+(a_1%2Ba_2)+\mod(c)|2425562|2425563|2425564|a_1+\mod(c)\cdot+a_2+\mod(c)\equiv+(a_1+\cdot+a_2)+\mod(c)|a_1+\mod(c)\cdot+a_2+\mod(c)\equiv+(a_1+\cdot+a_2)+\mod(c)|2425565|2425566|证明:让我们从左边进去，|2425567|2425568|a_1+\equiv+b_1+\mod(c)+,+++\:++a_2++\equiv+b_2+\mod(c)|a_1+\equiv+b_1+\mod(c)+,+++\:++a_2++\equiv+b_2+\mod(c)|2425569|2425570|我们可以写，|2425571|2425572|a_1%2Ba_2=b_1%2Bb_2%2B(k_1%2Bk_2)\cdot+c|a_1%2Ba_2=b_1%2Bb_2%2B(k_1%2Bk_2)\cdot+c|2425573|2425574|假设k_3=k_1%2Bk_2是一个新整数，|2425575|2425576|a_1%2Ba_2=b_1%2Bb_2%2Bk_3\cdot+c+\to+a_1%2Ba_2\equiv+b_1%2Bb_2+\mod(c)|a_1%2Ba_2=b_1%2Bb_2%2Bk_3\cdot+c+\to+a_1%2Ba_2\equiv+b_1%2Bb_2+\mod(c)|2425577|2425578|在哪里，|2425579|2425580|a_1+\mod(c)%2B+a_2+\mod(c)\equiv+(a_1%2Ba_2)+\mod(c)|a_1+\mod(c)%2B+a_2+\mod(c)\equiv+(a_1%2Ba_2)+\mod(c)|2425581|2425582|更简单地说，整数标量是可加的，您可以执行一次模算法。我们对乘法也有同样的证明方法。你可以取c的新标量，它是多个标量的乘法。|2425583|让我们按照简单的例子来做。28\equiv+2+\mod(26)和32+\equiv+6+\mod(26)。然后我们得到了28%2B32=60+\equiv+8+\mod(26)，当您首先计算模块并将它们相加时，它的结果是相同的。|2425584|我声明28+\cdot+32+\equiv+(2\cdot+6=12)+\mod(26)，而您可以通过预先乘以它们来检查它，然后找到剩余部分。因为28\cdot+32=+896=34\cdot+26%2B12。|2425585|总之，您不需要关心多个mods，只需先做手术，然后接受结果的模块。|2425586|此外，要了解更多细节，请查看此链接，https://brilliant.org/wiki/modular-arithmetic/|entityMap|0|INLINETEX|mutability|IMMUTABLE|k|1|\mod(26)|2|28|3|4|\{2,28,44,80,...\}|5|2%2Bk+\cdot+26|6|k_3=k_1%2Bk_2|7|28\equiv+2+\mod(26)|8|32+\equiv+6+\mod(26)|9|28%2B32=60+\equiv+8+\mod(26)|10|28+\cdot+32+\equiv+(2\cdot+6=12)+\mod(26)|11|28\cdot+32=+896=34\cdot+26%2B12|12|LINK|MUTABLE|url|https://brilliant.org/wiki/modular-arithmetic/^0|0|0|0|I|0|0|6|1|6|1|0|0|0|0|D|0|0|4|8|F|2|M|1|S|I|1F|C|4|8|1|F|2|2|M|1|3|S|I|4|1F|C|5|0|0|0|1C|0|0|0|0|1M|0|0|0|0|0|1I|0|0|0|0|0|W|0|0|2|B|2|B|6|0|0|0|1O|0|0|0|0|0|1C|0|0|0|D|J|X|K|1P|Q|D|J|7|X|K|8|1P|Q|9|0|3|15|21|T|3|15|A|21|T|B|0|0|I|1A|C^^$0|@$1|2|3|4|5|6|7|34|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|35|8|@]|9|@]|A|$]]|$1|C|3|D|5|E|7|36|8|@$F|37|G|38|H|I]]|9|@]|A|$J|-1|K|L]]|$1|M|3|-4|5|6|7|39|8|@]|9|@]|A|$]]|$1|N|3|O|5|6|7|3A|8|@$F|3B|G|3C|H|I]]|9|@$F|3D|G|3E|1|3F]]|A|$]]|$1|P|3|-4|5|6|7|3G|8|@]|9|@]|A|$]]|$1|Q|3|R|5|E|7|3H|8|@$F|3I|G|3J|H|I]]|9|@]|A|$J|-1|K|S]]|$1|T|3|-4|5|6|7|3K|8|@]|9|@]|A|$]]|$1|U|3|V|5|6|7|3L|8|@$F|3M|G|3N|H|I]|$F|3O|G|3P|H|I]|$F|3Q|G|3R|H|I]|$F|3S|G|3T|H|I]|$F|3U|G|3V|H|I]]|9|@$F|3W|G|3X|1|3Y]|$F|3Z|G|40|1|41]|$F|42|G|43|1|44]|$F|45|G|46|1|47]|$F|48|G|49|1|4A]]|A|$]]|$1|W|3|-4|5|6|7|4B|8|@]|9|@]|A|$]]|$1|X|3|Y|5|E|7|4C|8|@$F|4D|G|4E|H|I]]|9|@]|A|$J|-1|K|Z]]|$1|10|3|-4|5|6|7|4F|8|@]|9|@]|A|$]]|$1|11|3|-4|5|6|7|4G|8|@]|9|@]|A|$]]|$1|12|3|13|5|E|7|4H|8|@$F|4I|G|4J|H|I]]|9|@]|A|$J|-1|K|14]]|$1|15|3|-4|5|6|7|4K|8|@]|9|@]|A|$]]|$1|16|3|17|5|6|7|4L|8|@]|9|@]|A|$]]|$1|18|3|-4|5|6|7|4M|8|@]|9|@]|A|$]]|$1|19|3|1A|5|E|7|4N|8|@$F|4O|G|4P|H|I]]|9|@]|A|$J|-1|K|1B]]|$1|1C|3|-4|5|6|7|4Q|8|@]|9|@]|A|$]]|$1|1D|3|1E|5|6|7|4R|8|@]|9|@]|A|$]]|$1|1F|3|-4|5|6|7|4S|8|@]|9|@]|A|$]]|$1|1G|3|1H|5|E|7|4T|8|@$F|4U|G|4V|H|I]]|9|@]|A|$J|-1|K|1I]]|$1|1J|3|-4|5|6|7|4W|8|@]|9|@]|A|$]]|$1|1K|3|1L|5|6|7|4X|8|@$F|4Y|G|4Z|H|I]]|9|@$F|50|G|51|1|52]]|A|$]]|$1|1M|3|-4|5|6|7|53|8|@]|9|@]|A|$]]|$1|1N|3|1O|5|E|7|54|8|@$F|55|G|56|H|I]]|9|@]|A|$J|-1|K|1P]]|$1|1Q|3|-4|5|6|7|57|8|@]|9|@]|A|$]]|$1|1R|3|1S|5|6|7|58|8|@]|9|@]|A|$]]|$1|1T|3|-4|5|6|7|59|8|@]|9|@]|A|$]]|$1|1U|3|1V|5|E|7|5A|8|@$F|5B|G|5C|H|I]]|9|@]|A|$J|-1|K|1W]]|$1|1X|3|-4|5|6|7|5D|8|@]|9|@]|A|$]]|$1|1Y|3|1Z|5|6|7|5E|8|@]|9|@]|A|$]]|$1|20|3|21|5|6|7|5F|8|@$F|5G|G|5H|H|I]|$F|5I|G|5J|H|I]|$F|5K|G|5L|H|I]]|9|@$F|5M|G|5N|1|5O]|$F|5P|G|5Q|1|5R]|$F|5S|G|5T|1|5U]]|A|$]]|$1|22|3|23|5|6|7|5V|8|@$F|5W|G|5X|H|I]|$F|5Y|G|5Z|H|I]]|9|@$F|60|G|61|1|62]|$F|63|G|64|1|65]]|A|$]]|$1|24|3|25|5|6|7|66|8|@]|9|@]|A|$]]|$1|26|3|27|5|6|7|67|8|@]|9|@$F|68|G|69|1|6A]]|A|$]]]|28|$29|$5|2A|2B|2C|A|$K|2D]]|2E|$5|2A|2B|2C|A|$K|2F]]|2G|$5|2A|2B|2C|A|$K|2H]]|2I|$5|2A|2B|2C|A|$K|2G]]|2J|$5|2A|2B|2C|A|$K|2K]]|2L|$5|2A|2B|2C|A|$K|2M]]|2N|$5|2A|2B|2C|A|$K|2O]]|2P|$5|2A|2B|2C|A|$K|2Q]]|2R|$5|2A|2B|2C|A|$K|2S]]|2T|$5|2A|2B|2C|A|$K|2U]]|2V|$5|2A|2B|2C|A|$K|2W]]|2X|$5|2A|2B|2C|A|$K|2Y]]|2Z|$5|30|2B|31|A|$32|33]]]]

I try to explain it with more details. Suppose we have the following modular equation.
$$a \equiv b \mod(c)$$
Then for an integer $k$ we have,
$$a=b+k \cdot c$$
For example, in $\mod(26)$ the reminder for $28$ is equal to $2$ and also it is the same reminder for all $2+k \cdot 26$ including $\{2,28,44,80,...\}$. 
At this point we need to figure out what is the case of addition and multiplication. So I claim,
$$a_1 \mod(c)+ a_2 \mod(c)\equiv (a_1+a_2) \mod(c)$$
$$a_1 \mod(c)\cdot a_2 \mod(c)\equiv (a_1 \cdot a_2) \mod(c)$$
Proof: Let in the left side we have,
$$a_1 \equiv b_1 \mod(c) , \: a_2 \equiv b_2 \mod(c)$$ 
We can write,
$$a_1+a_2=b_1+b_2+(k_1+k_2)\cdot c$$
Assume $k_3=k_1+k_2$ as a new integer,
$$a_1+a_2=b_1+b_2+k_3\cdot c \to a_1+a_2\equiv b_1+b_2 \mod(c)$$
Whence,
$$a_1 \mod(c)+ a_2 \mod(c)\equiv (a_1+a_2) \mod(c)$$
More simply, the integer scalar is additive and you can execute the modular arithmetic once.
We have the same way of proof for the multiplication. You can take the new scalar of c is the multiplication of multiple ones.

Let follow the simple example. $28\equiv 2 \mod(26)$ and also $32 \equiv 6 \mod(26)$. Then we have $28+32=60 \equiv 8 \mod(26)$ while it has the same result when you first calculate the modulu and then add them up.

I claim $28 \cdot 32 \equiv (2\cdot 6=12) \mod(26)$ while you can check it by multiplying them in advance and then find the remainder. Because $28\cdot 32= 896=34\cdot 26+12$. 

As a conclusion, you don't need to care multiple mods and just do the operation first and then take the mod of the result.

Moreover, for more details have a look at this link,
<a href="https://brilliant.org/wiki/modular-arithmetic/" rel="nofollow noreferrer">https://brilliant.org/wiki/modular-arithmetic/</a>

I should prove mathematically that:
<blockquote>
$D_k(E_k(p))=p$
</blockquote>
I know
<blockquote>
$E_k(p)=c=(p+k)\bmod26$
</blockquote>
and
<blockquote>
$D_k(c)=p=(c-k)\bmod26$
</blockquote>
Substituting the two formulas to the initial assertion I have
<blockquote>
$((p+k)\bmod26-k)\bmod26=p$
</blockquote>
And applying the $\bmod$ property twice I have
<blockquote>
$((p \bmod26+k \bmod26)\bmod26)-k)\bmod26=p$
$((p\bmod26+k\bmod26)\bmod26)\bmod26-k\bmod26)\bmod26=p$
</blockquote>
But from here? How can I get rid of all that bunch of $\bmod$ to have $p$?

Prove monoalphabetic cipher is a valid symmetric cipher

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我应该从数学上证明：D_k(E_k(p))=p我知道呀E_k(p)=c=(p+k)\bmod26和D_k(c)=p=(c-k)\bmod26将这两个公式替换为我具有的初始断言((p+k)\bmod26-k)\bmod26=p并两次应用\bmod属性((p \bmod26+k \bmod26)\bmod26)-k)\bmod26=p ((p\bmod26+k\bmod26)\bmod26)\bmod

证明单字母密码是有效的对称密码-腾讯云开发者社区-腾讯云

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