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Solve $$\exists_k \mbox{ } 1001\cdot k+113 = x^{118} $$

How to deal when I have something like in that exercise when my big number is $ 13\cdot 11 \cdot 7$?
My current way: It is equivalent to: $$ x^{118} \equiv 1 \mod 7 \wedge x^{118} \equiv 3 \mod 11 \wedge x^{118} \equiv 9 \mod 13 $$

then I consider first equation: $$x^{118} \equiv 1 \implies x \equiv \pm1 \mod 7 $$ so $$ x \equiv 1 \vee x \equiv 6 \mod 7 $$

In first case I have that $$ x= 1 + 7k $$ but when I put that to $x^{118}$ I am not able to continue computing...

Bill Dubuque
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trolley
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1 Answers1

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Hint: By Fermat's little theorem, $x^6\cong1\pmod7\implies x^{118}\cong x^4\pmod7\implies x^4\cong1\pmod 7\implies x\cong1\lor x\cong6\pmod7$.

Now try to solve the other two congruences, and use the Chinese remainder theorem.

That is, by Fermat's little theorem again, $x^{10}\cong1\pmod{11}\implies x^{118}\cong x^8\pmod{11}\implies x^8\cong3\pmod{11}$ For instance, $2$ is a solution.

The CRT gives $57$ as a solution when we take $x\cong1\pmod7$ and $x\cong2\pmod{11}$. Now you have to apply it again with a solution to the third congruence. For instance, $4$ is a solution to the third congruence, and we get $134+1001k$ as one solution.

As far as I can see you will wind up with various possibilities for the combinations of the congruences, and will need to apply CRT in each case to finish. There is still some work to be done.

  • when you write $x^8\equiv3\pmod{11}$ and you say $x \equiv 2 \pmod{11}$ you checked in your had all numbers from $0$ to $10$ or you have trick to do that? @Chris – trolley May 11 '19 at 12:28
  • I checked in my head. –  May 11 '19 at 12:33
  • Ok. Other solution is $9$ so if I correctly understood I should check all combinations from solution in each equality? Chmm this is a lot of work to do as for 1 task from short exam... – trolley May 11 '19 at 12:35
  • Right. It does seem like a good amount of work. –  May 11 '19 at 12:39
  • I am going to solve that and post result there. – trolley May 11 '19 at 12:39
  • Sounds good. $9\cong -2\pmod{11}$, so yes. You have to get good at CRT. –  May 11 '19 at 12:42
  • I made mistake when I made post. Task was "Find number of solutions and give one of them" - can that task be simpler that current one? – trolley May 11 '19 at 12:46
  • Yes. That's easier. I got you a solution. Now you just need to find how many solutions to the third congruence (and multiply by $4$). –  May 11 '19 at 12:52
  • third congruence has 2 solutions: $x \equiv 4 \vee x \equiv 9 \mod 13$. So in each equation we have two possibilities. We can choose $2^3$ combinations of this possibilities. If all of them give me a solution (exactly one) then I have $8$ solution. But have I certainty that each triple give me exactly one solution, not less, not more? – trolley May 11 '19 at 12:59
  • I believe so. Each should give one solution mod $1001$. –  May 11 '19 at 13:01
  • Why should each give, can you explain me that? – trolley May 11 '19 at 13:02
  • By the CRT, the solutions are guaranteed. –  May 11 '19 at 13:03
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    @trolley To solve $,\Large x^{\Large 8}\equiv 3\pmod{!11},\ \rm\color{#c00}{\rm invert},$ it as follows $$\Large \bmod 11!:,\ x\not\equiv 0,\Rightarrow, \underbrace{x^{\Large 10}\equiv 1}{\rm Fermat},\Rightarrow,x^{\Large 2}\equiv\underbrace{ x^{\Large -8}\equiv \dfrac{1}3}{\rm\color{#c00}{inverted}}\equiv \dfrac{12}3\equiv 4,\Rightarrow, x\equiv \pm2\qquad\qquad $$ – Bill Dubuque May 12 '19 at 00:07
  • @trolley You can greatly reduce the effort needed computing all 8 combinations by first solving the CRT system for symbolic (variable) values, then specialize that for each of the $8$ cases - which is easy and quick. See my comment on your question for a few worked examples using this generic method. – Bill Dubuque May 12 '19 at 00:14