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The proof is given below:

enter image description here

But I do not understand the statement in the line before last which says "These give rise to the single congruence $a ^{560} \equiv 1 \pmod n$ where gcd(a, 561) = 1 ", I do not know why the previous mentioned system of linear congruences leads to this single congruence, what theorem is used to say this? is it the Chinese Remainder theorem ? if so how?

Intuition
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    Please don't post pictures: type the material. Pictures are not searchable, and can cause problems with sundry interfaces. – Arturo Magidin May 22 '19 at 21:29
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    $a^{560}-1$ is divisible by $3$, by $11$ and by $17$. – Angina Seng May 22 '19 at 21:29
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    Yes, it's the Chinese Remainder Theorem. The 3 congruences yield a unique solution modulo $3\times 11\times 17$; setting $x=a^{560}$, you get the unique solution $x\equiv 1\pmod{561}$ (because clearly that is a solution to the system, and by the CRT, it's the unique solution to the system modulo $561$). – Arturo Magidin May 22 '19 at 21:30
  • @ArturoMagidin That is just what I was going to put. – Mark Bennet May 22 '19 at 21:31
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    You can use properties of LCM, Euclid's Lemma, unique factorization, or CCRT = Constant-case CRT to make that inference - see here which explains all four methods. – Bill Dubuque May 22 '19 at 21:41
  • Ok I will follow this as long as I have enough time ..... sorry about this @ArturoMagidin – Intuition May 22 '19 at 22:14

4 Answers4

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The previous equations say that the integers $3,11$ and $17$ all divide $a^{560}-1$. This implies that their lcm must divide $a^{560}-1$ as well. But since $3,11,17$ are relatively prime their lcm is just their product which is $561$. So $561$ divides $a^{560}-1$, which means $a^{560}\equiv 1$(mod $561$).

Mark
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  • I like that this answer does not require the use of CRT. However, stylistically the proof in the OP does assume the reader’s familiarity with CRT (one immediately sees that some congruence of this shape holds, and that the precise value of the RHS can be verified very easily with almost no computation). – Erick Wong May 22 '19 at 22:17
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Since $a^{560}$ is congruent to $1$ modulo $3,11,$ and $17$, you know that $a^{560}-1$ is divisible by $3,11,$ and $17$. Thus, since $3,11,$ and $17$ are mutually coprime, $a^{560}-1$ is divisible by $3\cdot 11\cdot 17=561$ and so $a^{560}$ is congruent to $1$ modulo $561$.

Franklin Pezzuti Dyer
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You can see this as a consequence of the Chinese Remainder Theorem. Setting $x=a^{560}$, you have a system that says $$\begin{align*} x&\equiv 1\pmod{3}\\ x&\equiv 1\pmod{11}\\ x&\equiv 1\pmod{17} \end{align*}$$ By the Chinese Remainder Theorem, there is a unique solution modulo $3\times 11\times 17$, that is, $x\equiv b\pmod{561}$. But clearly, $b=1$ is a solution, so that is the unique solution. Hence $a^{560}=x\equiv 1\pmod{561}$.

Arturo Magidin
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It is indeed the chinese remainder theorem:

Chinese remainder Theorem

If a system of equations

$x \equiv a_i \mod m_i$ for some $i=1,2,\ldots,n$, with pairwise coprime $m_i$, then there is a unique solution mod $m_1m_2\cdots m_n$

Applying this to $m_1=3,m_2=11,m_3=17$ and $a_1=a_2=a_3=1$ gives the desired result.

Community
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auscrypt
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