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The context: I want to show how prime factorisation patterns repeat along the number line. I've taken three consecutive integers, the first containing 2^3, the second 3^2, the third 5^1. Taking respective multiples p, q, r, this gives me a set of equations and a set of conditions, namely neither 2 nor 3 nor 5 can divide any of them. I've also found that p, q, r must all end in a '1'. What's the most efficient way to search for a solution set other than (1, 1, 1)?,

Bill Dubuque
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Paul Stephenson
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  • Write down all solutions for the linear Diophantine equation $-8p+9q=1$, and then solve $9q+1=10r$. Then you have infinitely many solutions, e.g. $(p,q,r)=(91,81,73)$. Indeed, then $8p,9q,10r$ are consecutive integers, namely $728,729,730$. – Dietrich Burde Jun 25 '22 at 08:31
  • Hint: $ $ let $,\bar p = p-1,,\ \bar q = q-1,,\ \bar r = r-1.,$ Then

    $$\begin{align} &1+8p = 9q, = 10r-1\[.2em] \iff &\ \ \ \ \ \ \ 8\bar p = 9\bar q = 10\bar r =: ,m\[.2em] \iff &\ \ 8,9,10\mid m\[.2em] \iff &\ \ m = 360n,,\ {\rm by,\ lcm}(8,9,10)=360\[.2em] \iff &\ \ (\bar p,\bar q,\bar r) = (45n,40n,36n) \end{align}\qquad\qquad$$

     – Bill Dubuque Jun 25 '22 at 09:30
  • So the decrementing shift reduces it to the lcm universal property: $,a,b,c\mid m\iff [a,b,c]\mid m.,$ To calculate the lcm note $,[9,[8,10]] = [9,2[4,5]] = [9,40] = 360,$ where we used basic lcm properties: $,(a,b)=1\Rightarrow [a,b]=ab,, $ and lcm is associative, and the lcm distributive law. Or we can compute the lcm via prime factorizations – Bill Dubuque Jun 25 '22 at 10:15
  • Note that I require the exponents concerned to be no bigger than the ones I gave, hence my condition that 2, 3, 5 must divide none of p, q, r. – Paul Stephenson Jun 26 '22 at 09:41
  • $p,\color{#c00}q,r = 1!+!45n,,1!+!\color{#c00}{40}n,,1!+!36n,$ are all coprime to $30!\iff! 1!+!n,$ is coprime to $30,,$ e.g. $,\color{#c00}q = 1!+!\color{#0a0}{10}(4n),$ is coprime to $,\color{#0a0}{10},,$ and it is coprime to $3!\iff! 1=(3,1!+!\color{#c00}{40}n) = (3,1!+!\color{#c00}{1}n),,$ by $,\color{#c00}{40\equiv 1}\pmod{!3}\ \ $ – Bill Dubuque Jun 26 '22 at 11:11

1 Answers1

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If $9|(8p+1)$ then you can prove that $ 8p\equiv -1(mod9)$ so $p\equiv 1 (mod9)$.

Similarly if $10|(9q+1)$ then try and prove that $q\equiv 1(mod 10)$.

If $10|(8p+2)$ $\implies 5|(4p+1)$ then $p\equiv 1(mod 5)$. Since 9, 5 are coprime, you can write that $p\equiv 1(mod 45)$. Hence, $$p=45x+1, q=40x+1, r=36x+1$$ form a solution for natural x. I derived q and r via $q=\frac{8p+1}{9}$ and $r=\frac{9q+1}{10}$.

insipidintegrator
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  • Incorrect. Easier: by a shift it immediately reduces to an $,\rm lcm$ - see my prior comment on the question. – Bill Dubuque Jun 25 '22 at 09:56
  • @BillDubuque which part is wrong exactly? I haven’t done number theory for about two years so this answer was kind of a guess. Can you please point out the line which I should correct? – insipidintegrator Jun 25 '22 at 11:10
  • Your $nx$'s should be half that, e.g. $,p = 45x+1,$ for the general solution. Maybe you took the moduli product instead of their lcm. – Bill Dubuque Jun 25 '22 at 11:31
  • Isn’t the lcm of 9 and 10 equal to 90? @BillDubuque – insipidintegrator Jun 25 '22 at 13:06
  • You didn't $\color{#c00}{\rm cancel\ 2}$ here: $,10\mid 8p!+!2!!!\overset{\large \color{#c00}{\div 2!}}\iff 5\mid 4p!+!1\iff p\equiv 1\pmod{!5},,$ so your claimed solution $,p\equiv 1\pmod{!10},$ is missing the solutions $,p\equiv 6\pmod{!10},,$ e.g. $,p=6.\ \ $ – Bill Dubuque Jun 25 '22 at 13:42
  • Ah I see. Thanks @BillDubuque – insipidintegrator Jun 25 '22 at 14:04