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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.

Bob Dobbs
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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

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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 $\,1 \!=\! (b,b') = (a',b'),\,$ 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 $\,f(x)\!=\!(\color{#0a0}{ax\!+\!1})(\color{#c00}{bx\!+\!1})\,$ has a root $\!\color{#0a0}{\bmod a'}\,$ & $\!\color{#c00}{\bmod b'},\:\!$ which CRT lifts to a root $\!\bmod \color{#0a0}{a'}\color{#c00}{b'}\!=\!n$.

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