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I have a feeling that induction might be necessary since the question includes "for every integer $n ≥ 2$".

So with this in mind, the base case would be $n=2$. If $(2x+1)(3x+1)\equiv0\pmod 2$, then $x=1$ is a solution because $12\equiv 0\pmod 2$ is a true statement.

For the inductive step, we assume that $$(2x+1)(3x+1)\equiv0\pmod{k}$$ and we need to show that $$(2x+1)(3x+1)\equiv 0\pmod{ (k+1)}.$$ I'm unsure what to do from here.

Bill Dubuque
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FoiledIt24
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  • Please see https://math.meta.stackexchange.com/questions/5020/ – Angina Seng Jul 02 '19 at 01:40
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    Chinese remainder theorem? – Angina Seng Jul 02 '19 at 01:44
  • We can use strong induction, base case is what you showed. For inductive step, we have $k+1=p_1^{r_1}p_2^{r_2}...$ where the $p_i's$ are distinct primes. Apply induction on each of the $p_i^{r_i}$ to get $(2x+1)(3x+1)≡0(mod p_1^{r_1})$, $(2x+1)(3x+1)≡0(mod p_2^{r_2})$ and so on. Since each of the $p_i^{r_i}$ are coprime, we can apply Chinese remainder theorem to conclude that $(2x+1)(3x+1)≡0(mod k+1)$ has a solution! – user 6663629 Jul 02 '19 at 01:56
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    If $n$ is odd, then $x=\dfrac{n-1}2$ is a solution; if $n$ is not a multiple of $3$ then $n-1$ or $2n-1$ is a multiple of $3$ so $\dfrac{n-1}3$ or $\dfrac{2n-1}3$ is a solution – J. W. Tanner Jul 02 '19 at 02:01

3 Answers3

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Theorem $\ f(x)= (ax\!+\!1)(bx\!+\!1)\ $ has a root $\!\bmod n\,$ for all $\,n\!>\!1\!\iff\! (a,b) = 1$

Proof $\ (\Rightarrow)\ $ if $\,1 < c\mid a,b\,$ then $\!\bmod c\!: f(x)\equiv 1\not\equiv 0.\,$ $(\Leftarrow)\,$ Factor $\,n = a'b'$ where $\,b'\,$ collects all prime factors of $\,n\,$ that divide $\,a,\,$ so $\,1 \!=\!{(a,a')},\,$ and $\,1 \!=\! (b,b') = \color{darkorange}{(a',b')},\,$ by $\,(a,b)\!=\!1,\,$ so $\,\color{#0a0}{ax\!+\!1}\,$ has root $\,\color{#0a0}{x\equiv -a^{-1}\!\pmod{\!a'}}\,$ and $\,\color{#c00}{bx\!+\!1}\,$ has root $\,\color{#c00}{x\equiv -b^{-1}\!\pmod{\!b'}}.\,$ Therefore $\,(\color{#0a0}{ax\!+\!1})(\color{#c00}{bx\!+\!1})\,$ has a root $\!\color{#0a0}{\bmod a'}\,$ & $\!\color{#c00}{\bmod b'},\:\!$ so CRT lifts it to $\!\bmod \color{#0a0}{a'}\color{#c00}{b'}\!=\!n,\,$ by $ \color{darkorange}{(a',b')}\!=\!1$.

Remark $ $ See this MO post for more on local-global principles / obstructions (Hasse, Selmer, etc).

Bill Dubuque
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Here is a solution without induction on $n$. Let $n=2^am$ with $m$ odd. Then $2^a$ is not a multiple of $3,$

so $2^a-1$ or $2^{a+1}-1$ is a multiple of $3, $ and $x=\dfrac{2^a-1}3$ or $x=\dfrac{2^{a+1}-1}3 $ is a solution$\mod 2^a$.

Also, $x=\dfrac{m-1}2$ is a solution $\mod m$.

Therefore, by the Chinese remainder theorem, there is a solution $\mod n$.

J. W. Tanner
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Without CRT but EL.

Let $n=2^km$ where $m$ is odd.

Since $\gcd(2,m)=1$ $\exists t_1,t_2\in\Bbb Z$ such that $2t_1+1=t_2m$.

Since $\gcd(3,2^k)=1$ $\exists s_1,s_2\in\Bbb Z$ such that $3s_1+1=s_22^k$.

Since $\gcd(m,2^k)=1$ $\exists u,v\in\Bbb Z$ such that $um+v2^k=1$.

Let $x=t_1+(s_1-t_1)um.$

Then $(2x+1)(3x+1)\equiv 0\pmod n.$

Bob Dobbs
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  • @BillDubuque But I didn't use CRT directly. Right? – Bob Dobbs Dec 09 '23 at 18:01
  • @BillDubuque Omg. You wrote an his tory on that subject. – Bob Dobbs Dec 09 '23 at 19:11
  • You do apply CRT (without saying its name). Your first three equations are equivalent to $,\color{#0a0}{t_1\equiv -1/2}\pmod{!m},,$ $\color{#0a0}{s_1\equiv -1/3}\pmod{!2^k},,$ $\color{#c00}{u\equiv m^{-1}}!\pmod{!2^k}.,$ Then you claim

    $$\begin{align}&x\equiv \color{#0a0}{t_1}!!!\pmod{m}\&x\equiv \color{#0a0}{s_1}!!!\pmod{2^k}\end{align}\Longleftarrow\ x\equiv t_1 + (s_1-t_1),m(\underbrace{\color{#c00}{m^{-1}}\bmod 2^k}_{\textstyle \color{#c00}u^{\phantom{\tiny i}}})\qquad$$ which is exactly the (Easy) CRT formula.$\ \ $

     – Bill Dubuque Dec 09 '23 at 20:47
  • Thus what you wrote is exactly the same method I used (but I didn't apply any specific CRT formula since here we need only the existence of a root). So it is a duplicate of the method in my answer. It's best not to duplicate answers unless there is a very good reason to do so. $\ \ $ – Bill Dubuque Dec 09 '23 at 20:47
  • @BillDubuque For this particular question it is obviously "another way" solution. – Bob Dobbs Dec 09 '23 at 21:19
  • No, it is exactly the same way, as I explained above. But doing it as above obfuscates the use of CRT, so obscures the underlying key idea. – Bill Dubuque Dec 09 '23 at 21:27
  • I started from EDL which is more basic. I am not trying to degrade CRT or other answers – Bob Dobbs Dec 10 '23 at 06:33
  • The point is that you duplicated a common (EDL-based) proof of the (Easy) CRT formula. But the proof does not simplify for this special case, so it is better to simply invoke CRT by name (rather than by value, i.e. repeating its proof). Moreover, we don't need the solution. Rather we need only the existence of a solution to the congruence system - which follows from the CRT solvability criterion, i.e. the moduli are coprime. Removing this unneeded info yields the proof in my prior answer (in OP special case). $\ \ $ – Bill Dubuque May 21 '24 at 16:09
  • What should I do know? Should İ delete this unnecessary answer? @BillDubuque – Bob Dobbs May 21 '24 at 17:47