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$$x^3\equiv x \pmod{105}$$

I'm trying to solve this equation. Here's what I tried so far:

$$x^3\equiv x \pmod{105} \iff x^2\equiv 1 \pmod{105}$$

Then, applying the Chinese remainder theorem, I got the system: $$\cases{x^2 \equiv 1 \pmod{5}\\x^2 \equiv 1 \pmod{7}\\x^2 \equiv 1 \pmod{3}}$$ With the following solutions:

$$\cases{x \equiv \pm1 \pmod{5}\\x \equiv \pm1 \pmod{7}\\x \equiv \pm1 \pmod{3}}$$ At this point, I think I need to pretty much solve these eight systems:

$$\cases{x \equiv 1 \pmod{5}\\x \equiv 1 \pmod{7}\\x \equiv 1 \pmod{3}} \cases{x \equiv 1 \pmod{5}\\x \equiv 1 \pmod{7}\\x \equiv -1 \pmod{3}} \cases{x \equiv 1 \pmod{5}\\x \equiv -1 \pmod{7}\\x \equiv 1 \pmod{3}} \cases{x \equiv -1 \pmod{5}\\x \equiv 1 \pmod{7}\\x \equiv 1 \pmod{3}}$$$$ \cases{x \equiv -1 \pmod{5}\\x \equiv 1 \pmod{7}\\x \equiv -1 \pmod{3}} \cases{x \equiv -1 \pmod{5}\\x \equiv -1 \pmod{7}\\x \equiv 1 \pmod{3}} \cases{x \equiv 1 \pmod{5}\\x \equiv -1 \pmod{7}\\x \equiv -1 \pmod{3}} \cases{x \equiv -1 \pmod{5}\\x \equiv -1 \pmod{7}\\x \equiv -1 \pmod{3}}$$

Here's how I solved the first one: Considering the first two equations, we get: $$x=5k+1=7h+1$$ from which $k = 7+7y, h = 5+5y$, with $y \in \mathbb{Z}$. Therefore, $$x=36+35y\iff x\equiv1\pmod{35}$$ Adding in the third equation, we have that $36+35y = 1+3 w$, from which $x = 1281 + 35w \iff x \equiv1\pmod{105}$.

However, this one seems like a really tedious method as I'd have to do the same calculations for seven more systems. Is there anything I'm missing? Is there a faster way to do this?

Martin Sleziak
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Samuele B.
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    Don’t forget that zero is a solution too. – Michael Burr Apr 13 '20 at 11:05
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    $x^3\equiv x \pmod{105} \iff x^2\equiv 1 \pmod{105}$ is not entirely true. – Arthur Apr 13 '20 at 11:09
  • If you use the general formula for the CRT system, the $\pm 1$ are just different coefficients to plug in at the end, so just 8 small calculations. E.g. see here. – Henno Brandsma Apr 13 '20 at 11:12
  • $x^{3}-x=(x-1)x(x+1)$ is a multiplication of 3 consecutive integers. This is always divisible by 3 so don't worry about that. Try to find pattern in which multiple of 5 and 7 are both within 3 consecutive integers – acat3 Apr 13 '20 at 11:21
  • Tricks to cut down work: 1) half of your answers are negatives of others. This means you only need to solve four. 2) If you compute iteratively, you can reuse common steps for pairs that have the same values. 3) If $x\equiv a\mod r$ and $\equiv a\mod s$, then $x\equiv a\mod rs$. – Michael Burr Apr 13 '20 at 11:25
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    the first (and last) case is the constant case of the Chinese remainder theorem, so it doesn't have to be so tedious; if $x\equiv1\bmod 3,5, $ and $7$, then $x\equiv 1\bmod 105$; likewise, if $x\equiv-1\bmod3,5,$ and $7$, then $x\equiv-1\equiv104\bmod105$ – J. W. Tanner Apr 13 '20 at 14:20
  • I added an answer showing the method using the generic (symbolic) CRT formula. See the link there for many more worked examples (with further detail and optimizations). – Bill Dubuque Apr 13 '20 at 22:14

2 Answers2

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Hint $ $ It's always true mod $3,\,$ so by CRT we need only combine all roots $\{0,\pm1\}$ mod $5$ and $7,\,$ and $\,x\equiv a\pmod{\!5},\,x\equiv b\pmod{\!7}\!\iff\! x\equiv b+14(b-a)\pmod{\!35}.\,$ For $\,a,b\in \{0,\pm1\}$ this yields $\,x\equiv \pm \{0,1,6,14,15\}\pmod{\!35}$

Bill Dubuque
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As $x^3-x=(x-1)x(x+1)$ is a product of three consecutive integers

$3$ must divide $x^3-x$

So, we need $$x^3\equiv x\pmod{5\cdot7}$$

If $(x-1)x(x+1)\equiv0\pmod 5$

$\implies x\equiv0\ \ \ \ (1), x\equiv-1\ \ \ \ (2), x\equiv1\pmod5\ \ \ \ (3)$

Similarly, $x\equiv0\ \ \ \ (4), x\equiv-1\ \ \ \ (5), x\equiv1\pmod7\ \ \ \ (6)$

Now apply CRT on $(1),(4); (1),(5);(1),(6);(2),(4); (2),(5);(2),(6)$

lab bhattacharjee
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  • Cases (1)(4); (2)(5); (3)(6) can be solved without CRT. For the remaining 6 cases, half of them are negatives of the other, so CRT only needs to be used three times. – Michael Burr Apr 13 '20 at 11:58