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Show that for each natural number $n≥2,$ there exists an integer $a>0$ such that $(6a+1)a$ leaves a remainder of $1$ upon division by $n.$

This question was on the Irish Team Selection Test for this year.

The question comes down to solving the congruence $6a^2 + a - 1\equiv_n0.$ We may factor this and write it as $(3a-1)(2a+1)\equiv_n0.$ So, if $n$ is odd, it suffices to take $a=\frac{n-1}{2}.$ If $n$ is $1$ mod $3,$ then $2n$ is $2$ mod $3,$ and hence we may take $a=\frac{2n+1}{3}.$ If $n$ is $2$ mod $3,$ take $a=\frac{n+1}{3}.$

The only case we are left with is when $n$ is a multiple of $6.$ I do not know how to proceed in this case. One thing I noticed is that in this case, taking $a=n-5$ works many times for small values of $n.$ This happens only for those $n$ that divide $144$ (and are multiples of $6,$ of course). It will be clear why this happens once $6a^2+a-1$ is expanded with $a=n-5.$ Beyond this, I have no idea.

How to deal with this case?

aqualubix
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    When $n=6c$ , then $a=6c+1$ will work ! You have already solved all other Cases ! – Prem Dec 09 '23 at 08:58
  • @Prem, I am not sure that that works. Let $n=30.$ Then, $a=31$ gives $(6a+1)a\equiv_{30}7.$ – aqualubix Dec 09 '23 at 09:19
  • If you can show it for n = power of 2 and power of 3 then probably by Chinese remainder theorem you can show it for power of 6 – Tony Pizza Dec 09 '23 at 09:29
  • When $n=6c,a=6b+1$, then $(9b+1)(4b+1)$ must be a multiple of $c$ – Empy2 Dec 09 '23 at 09:39
  • Let $,n = 2^k m,\ 2\nmid m.,$ CRT $\Rightarrow \exists,a!:$ $,a\equiv \frac{1}3\pmod{!2^k},,$ $,a\equiv -\frac{1}2\pmod{!m},,$ so $,2^k\mid 3a!-!1,,$ $,m\mid 2a!+!1,,$ so $\ n=2^k m\mid (3a!-!1)(2a!+!1).\ $ This is a duplicate. $\ \ $ – Bill Dubuque Dec 09 '23 at 09:49
  • Finding a dupe target turned out to be easier than I thought it would be. – Bill Dubuque Dec 09 '23 at 09:58
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    You are right , @aqualubix , that will not work. The Dup gives 3 alternative Solutions. – Prem Dec 09 '23 at 17:23

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