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There is a system of Diophantine equations: \begin{equation*} \begin{cases} 368=x^7 (mod 407)\\ 389=x^{11}(mod 407) \end{cases} \end{equation*}

However, solving each of them by hand is quite a difficult task. The question is: how knowing both of the equations ease the task? How could they be solved manually in that case?

SilverLight
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  • Have you tried the Chinese remainder theorem? And your question is unclear to me; do you want to know how to solve the system of equations manually? Or do you want to know how to solve the system manually given the solutions to each of the individual equations? – Servaes Apr 13 '19 at 16:40
  • @Servaes Sorry, I've changed the question - now it should be clear. I want to know how to solve the system manually. Talking about the Chinese remainder theorem, I've thought it's used for linear systems, isn't it? – SilverLight Apr 13 '19 at 16:51
  • Hint: $\gcd(7,11) = 1,$ so you can get the exponent $1$ on $x$ by a linear combination of $7$ and $11$ – Bill Dubuque Apr 13 '19 at 17:00

3 Answers3

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Here's one way.$$\begin{align} &x^{21}\equiv368^3\equiv103\pmod{407}\implies x^{22}\equiv103x\pmod{407}\\ &x^{22}\equiv389^2\equiv324\pmod{407}\end{align}$$

So we just have to solve $$103x\equiv324\pmod{407}$$

saulspatz
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  • Thanks! It's completely clear! – SilverLight Apr 13 '19 at 17:12
  • @SilverLight You're welcome. I should have mentioned that we need to check that the answer is actually a solution to both equations. The argument above shows only that $15$ is the only possible solution, not that it actually is one. – saulspatz Apr 13 '19 at 17:14
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The Bezout equation $\gcd(7,11)=\underbrace{\color{#f84}{\bf 1} =\color{#0a0}{11\cdot 2} - \color{#c00}{7\cdot 3}}\,$ yields exponent $\color{#f84}{\bf 1}$ on $\,x\,$ as follows

$$\qquad\qquad\qquad\qquad\ \ x^{\Large\color{#f84}{\bf 1}}\equiv\, \dfrac{(x^{\Large \color{#0a0}{11}})^{\Large \color{#0a0}2}}{(x^{\Large \color{#c00}{7}})^{\Large \color{#c00}3}}\equiv \dfrac{389^{\large 2}}{368^{\large 3}}\equiv \dfrac{-83}{103}\equiv 15\!\pmod{\!407} $$

i.e. $\ x\equiv -83\cdot 103^{-1}.\,$ The inverse can be computed by the Extended Euclidean algorithm.

Notice that our inferences are unidirectional, i.e. any solution of the system must satisfy the above. They do not imply that $\,x\equiv 15\,$ is a solution, so you need to check (or prove) that it is.

Bill Dubuque
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  • but how could i calculate $389^2/368^3$? It's something terrifying – SilverLight Apr 13 '19 at 17:14
  • @SilverLight It boils down to exactly the same (omitted) computation as in the other answer. So why did you accept it if you don't know how to finish? – Bill Dubuque Apr 13 '19 at 17:17
  • in the other answer there is a linear Diophantine equation, which i've just solved manually. – SilverLight Apr 13 '19 at 17:23
  • @SilverLight $, x\equiv b/a\equiv ba^{-1}\iff ax\equiv b,$ so it is the same linear congruence. I used fractions to make vivid that it is simply the Bezout equation done in the exponents. – Bill Dubuque Apr 13 '19 at 17:26
  • I see now, thanks! I'm just not very familiar with all that ring theory. – SilverLight Apr 13 '19 at 17:32
  • However, if $ gcd(a,b) \neq 1$, does it change anything? – SilverLight Apr 13 '19 at 17:34
  • @SilverLight $,a/b := ab^{-1}$ is well-defined $!\iff! b^{-1}$ exists $!\iff b ,$ is coprime to the modulus (by Bezout). If you are not familiar with modular fractions then simply replace $,a/b,$ by $,ab^{-1},$ above. Then we get $,x\equiv -83(103^{-1}),$ so $,103x\equiv -83\equiv 324,,$ same as the other answer. – Bill Dubuque Apr 13 '19 at 17:52
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To expand on my comment; because $407=11\times37$, by the Chinese remainder theorem the system is equivelent to \begin{eqnarray} x^7&\equiv&368\equiv5&\pmod{11},\tag{1}\\ x^7&\equiv&368\equiv35&\pmod{37},\tag{2}\\ x^{11}&\equiv&389\equiv4&\pmod{11},\tag{3}\\ x^{11}&\equiv&389\equiv19&\pmod{37}.\tag{4} \end{eqnarray} In particular $x\not\equiv0\pmod{11}$ and $x\not\equiv0\pmod{37}$, so by Fermat's little theorem $$x^{10}\equiv1\pmod{11} \qquad\text{ and }\qquad x^{36}\equiv1\pmod{37}.$$ Then congruence $(3)$ immediately shows that $x\equiv4\pmod{11}$, and congruences $(2)$ and $(3)$ together show that $x^{18}\equiv-1\pmod{37}$. Then a bit of fiddling around yields the identity $$x\equiv x^{37} \equiv-x^{37+18} \equiv-x^{11\times 5} \equiv -19^5\pmod{37},$$ and a quick computation then shows that $x\equiv15\pmod{37}$, so by the Chinese remainder theorem the solution to the system is $x\equiv15\pmod{407}$.

Servaes
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