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I am attempting to find a simple way to find solutions to $x^2$ - 6x + 8 = 0 mod 105

105 = 3 * 5 * 7, so I can do mod 3, mod 5, and mod 7 to 105 and see the overlap, but that takes a lot of time to list them out and find matches.

Also, we can factor the equation into (x-2)(x-4) = 0 mod 105, or $x^2$ - 6x + 8 - 105m = 0

Please let me know some methods I haven't realized yet. I used a calculator that told me 2,4, 32, 37, 44, 67, 74, and 79, but I think it went brute force rather than a method.

I don't have much understanding of CRT, so I'm looking for alternative methods

Knocker379
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  • The $(x-2)(x-4)\equiv 0\pmod{105}$ is the way to go. Either $x=2,4\pmod{105}$ or ... – TheBestMagician Jul 12 '21 at 23:25
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    You edited the question after closure to say you are not familiar with CRT. But there are no general methods that are simpler than CRT, so you should learn that basic result. It's very easy using CRT - see the linked dupes. Note if $r$ is a root then so too is $6-r$ so we only need to compute $4$ of the $8$ roots $,x\equiv (a,b,c)\bmod (3,5,7),,$ for $,a,b,c\in{2, 4}\ \ $ – Bill Dubuque Jul 13 '21 at 00:06
  • @TheBestMagician Or what? It is not clear what your ellipses denote. – Bill Dubuque Jul 13 '21 at 00:08
  • @BillDubuque It was a hint, the other option is if (for example), something like $x-2, x-4$ are factors mod 105 that multiply to 105 – TheBestMagician Jul 13 '21 at 00:09
  • @TheBestMagician Not "multiply to $105$" but, rather, $105n$, i.e. a multiple of $105.$ That's equivalent to the method I describe in the linked dupes. – Bill Dubuque Jul 13 '21 at 00:16
  • @BillDubuque I was speaking mod 105 – TheBestMagician Jul 13 '21 at 00:18
  • We can directly apply this answer in the 3rd dupe, which computes all the sqrts $r$ of $,4\bmod 105$. By the quadratic formula our roots are $, (6\pm r)/2\equiv 3\pm r/2,,$ e.g. the sqrt $,r\equiv 23,$ yields $,3+23/2\equiv 3+128/2\equiv 67.,$ Alternatively, apply the same method as is the remark, but replace the 2nd root $,-2,$ by $,4,,$ i.e. use the below Easy CRT formula $$\begin{align} x&\equiv 4!\pmod p\ x&\equiv 2!\pmod{qr}\end{align}!\iff x\equiv 2\ +\ qr\left[,\dfrac{2}{qr}\ \bmod\ p,\right]!!\pmod{pqr}\qquad$$ – Bill Dubuque Jul 13 '21 at 01:19
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    Oh, Chinese Remainder Theorem. I was wondering what critical race theory had to do with any of this. – John Douma Jul 13 '21 at 02:41

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