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I'd like to solve a system of modular equations, but it's a somewhat unusual system. Specifically, what is the smallest natural number $x$ for which $x = 2\pmod{6y+1}$, and $x=3\pmod{12y+1}$, and $1 \pmod{18y+1}$ for some natural number $y$?

If $y=1$, then WolframAlpha tells me that $x=1654$. But is it possible that a bigger choice of $y$ could yield a smaller value of $x$? This came up in the context of Godel coding, by the way.

Some Guy
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Keshav Srinivasan
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3 Answers3

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The bases, call them $p,q$ and $r$, are coprime, so try to solve $$aqr=2\pmod p\\ bpr=3\pmod q\\ cpq=1\pmod r$$ The answer is $aqr+bpr+cpq\pmod{pqr}$

Empy2
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    And how would I go about solving these things, considering that y appears both in the equation and what you're modding by? – Keshav Srinivasan Feb 14 '21 at 04:24
  • $qr=(12y+1)(18y+1)\pmod{6y+1}=(-1)(-2)\pmod{6y+1}$ – Empy2 Feb 14 '21 at 04:32
  • OK thanks, that makes sense. – Keshav Srinivasan Feb 14 '21 at 04:59
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    @Keshav Note that the above is a rote application of a common CRT formula (and you may get stuck in the more difficult omitted calculations of $b$ and $c)$. But the CRT result is not enough to answer your question - it yields $,x\equiv f(y)\pmod{(6y+1)(12y+1)(18y+1)},$ for a polynomial $,f(y),$ of degree $\le 2.,$ You seek the minimum natural value it can take (for all $,y,,$ or for fixed $,y,$?). This requires more work. – Bill Dubuque Feb 14 '21 at 10:28
  • If you don't know how to finish you should probably unaccept the answer so that others will see the question is not resolved, and they can help you do the more difficult steps. – Bill Dubuque Feb 14 '21 at 10:28
  • $4pr=(1)(-1)$, so $b4pr=-b=12\pmod q$...Yes, that was difficult... – Empy2 Feb 14 '21 at 11:43
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    I can assure you from much experience teaching these topics that scaling tricks like that often prove troublesome to many beginners. Also that is not a rigorous proof (you need to justify why scaling and cancelling $4$ is valid; answer: because $4$ is invertible mod $12y+1$ being coprime to it). – Bill Dubuque Feb 14 '21 at 18:44
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\begin{align} x &\equiv 2\pmod{ 6y+1} \\ x &\equiv 3\pmod{12y+1} \\ x &\equiv 1\pmod{18y+1} \end{align}

Note \begin{align} 2( 6y+1) - 1(12y+1) &= 1 \\ 3(12y+1) - 2(18y+1) &= 1 \\ -2(18y+1) + 3( 6y+1) &= 1 \\ \end{align}

So the numbers $6y+1, 12y+1$, and $18y+1$ are pairwise prime.

\begin{align} (12y+1)(18y+1) &\equiv 216y^2+30y+1 \pmod{6y+1} \\ &\equiv (216y^2+30y+1)\pmod{6y+1} \\ &\equiv (36y-1)(6y+1) + 2 \pmod{6y+1} \\ &\equiv 2 \pmod{6y+1} \\ &\equiv 0 \pmod{12y+1} \\ &\equiv 0 \pmod{18y+1} \end{align}

\begin{align} -12(6y+1)(18y+1) &\equiv (-1296y^2-288y-12) \pmod{12y+1} \\ &\equiv (-108y - 15)(12y + 1) + 3 \pmod{12y+1} \\ &\equiv 0 \pmod{6y+1} \\ &\equiv 3 \pmod{12y+1} \\ &\equiv 0 \pmod{18y+1} \end{align}

\begin{align} (9y+5)(6y+1)(12y+1) &\equiv 648y^3 + 522y^2 + 99y + 5 \\ &\equiv (36y^2 + 27y + 4)(18y + 1) + 1 \pmod{18y+1} \\ &\equiv 0 \pmod{6y+1} \\ &\equiv 0 \pmod{12y+1} \\ &\equiv 1 \pmod{18y+1} \end{align}

\begin{array}{c} (12y+1)(18y+1)-12(6y+1)(18y+1)+(9y+5)(6y+1)(12y+1) \\ = 648y^3-558y^2-159y-6 \end{array}

So, a solution is $$648y^3-558y^2-159y-6 \pmod{(6y+1)(12y+1)(18y+1)}$$

Steven Alexis Gregory
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I don't have reputation to comment, so trying to answer here.

To make all 3 modulo co-prime to each other. we take smallest y>0. naturally if we take y=1 then we have modulo as {7,13,19} and they are prime numbers. so, choice of y seems good.

  • Now the smallest number should be within ${7*13*19=1729}$, so 1654 is the smallest answer you get.
  • I don't know what could be the technique which gives higher value of y and smaller value of x.
SSA
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    "so, choice of y seems good." Seems to be doesn't mean it is. "I don't know what could be the technique which gives higher value of y and smaller value of x." Not knowing doesn't mean one doesn't exist. – fleablood Feb 14 '21 at 07:44