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I need to find all numbers

\begin{equation} 6x^2 + 2x \equiv 20 \pmod{513} \end{equation}

and I'm having a trouble since 513 is not prime number

I tried looking for roots but i end up with

\begin{equation} 2(x+2)(3x-5) \equiv 0 \pmod{513} \end{equation}

which lead to nothing in this case.

I also wanted to somehow find a square foirmual of my equation to make it a bit easier but I

failed doing it and now I'm running out off ideas how to solve it.Any help would be much

appreciated.

I found way to write it down as

\begin{equation} (6x+1)^2 \equiv 121 \pmod{513} \end{equation} and then

\begin{equation} 6x \equiv 6 \pmod{19}\\ 6x \equiv 11 \pmod{19}\\ 6x \equiv 12 \pmod{27}\\ 6x \equiv 13 \pmod{27}\\ \end{equation}

but after solving it I get bad answers so is my approach close to right one?

Bill Dubuque
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Karol Bargieł
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  • Can you explain how you found $2(x+2)(3x+5)\equiv 0$ and $(6x+1)^2\equiv 121$? They both seem suspicious to me. – Captain Lama Jan 20 '24 at 11:37
  • sure I multiplied both sides by 6 and got 36x^2+12x-120=0, then (36x^2+12x+1)-121=0, (6x+1)^2=121 – Karol Bargieł Jan 20 '24 at 11:45
  • Unfortunately you are not allowed to multiply by $6$ because $6$ is not invertible modulo $513$. Rather, you are (of course) allowed, but the equation you get is no longer equivalent to the original (you will get more solutions than the original equation). – Captain Lama Jan 20 '24 at 11:48
  • oh yes you're right ,but then right now I don't really know what to do in that case – Karol Bargieł Jan 20 '24 at 11:53
  • You can apply the CRT for $19$ and $27$. – Dietrich Burde Jan 20 '24 at 12:01
  • It is not quite over since the leading coefficient is not invertible mod $27$. – Captain Lama Jan 20 '24 at 12:02
  • I dont think so because there isn;t any alternate form to wrote it down as square – Karol Bargieł Jan 20 '24 at 12:08
  • You do have $x=5\pmod{3^3}$ and $x=9,13\pmod{19}$. Apply chinese residue theorem. – Piquito Jan 20 '24 at 12:12
  • Also, in your derivation of $(6x+1)^2\equiv 121$, there seems to be a mistake, you get $6x$ and not $12x$ in the quadratic expression, after multiplying by $6$. Unless there is a typo in the original equation? – Captain Lama Jan 20 '24 at 12:13
  • i made a mistake it should be 2x instead of one – Karol Bargieł Jan 20 '24 at 12:24
  • @Piquito how do you know that x=5 mod 27 and 9,13 are mod 19? – Karol Bargieł Jan 20 '24 at 13:35
  • The same CRT method in the dupe works here: simply CRT combine the obvious roots $,{−2}\bmod 27,$ with the roots $,{−}2,,8\bmod 19\ \ $ – Bill Dubuque Jan 20 '24 at 19:57
  • @Cap We can apply the quadratic formula if we also scale the modulus by $6$ (as explained in the 2nd linked dupe), so $\bmod 6\cdot 513!:\ x\equiv \dfrac{-1\pm\sqrt{11^2}}6,,$ with $,s = \sqrt{11^2}\equiv \pm11,\pm 961$ must be $\color{#c00}{{-}11,+961},$ by $,s\equiv 1\pmod{!3}$ for $3$ to cancel from the denom (so the fraction exists). We compute the nontrivial sqrt $,s=961$ by CRT solving $,s\equiv 11\pmod{!38},\ s\equiv -11\pmod{!81}.,$ Finally $\bmod 513!:\ x\equiv (-1+{\color{#c00}{-11,961}})/6 \equiv -2,160\ \ $ – Bill Dubuque Jan 20 '24 at 20:43

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