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I need help with the following question:

Find all the integer solutions for: $3x^2+18x+95\equiv 0\pmod {143}$

My solution: First I know that $143=11\cdot 13$ then because $\gcd (11,13) = 1$ then $3x^2+18x+95\equiv 0\pmod {143}$ if, and only if $$3x^2+18x+95\equiv 3x^2+7x+7\equiv 0\pmod {11} \\ 3x^2+18x+95 \equiv 3x^2+5x+4\equiv 0\pmod {13}$$

I don't know how to solve those equations and I don't know how to combine it to the big solution for the real question (I know about the CRT, but I didn't realy understood how to use it, I'd love help with this).

thanks in advance

Bill Dubuque
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The student
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    For combining, use the Chinese remainder theorem. For the two congruences, multiply with a suitable constant, then complete the square to get congruences of the form $(x+a)^2 \equiv b \pmod{p}$. – Daniel Fischer Jul 30 '20 at 21:50
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    You could solve by completing the square: $3x^2+7x+7\equiv0\implies x^2+6x+6\equiv0\implies$ $ (x+3)^2\equiv3\implies x+3\equiv\pm5\implies x\equiv2$ or $3\bmod11$ – J. W. Tanner Jul 30 '20 at 21:52
  • $3x^2+18x+95-143=0$ has solutions, x=2 and-8, so it may help by thinking about $3(2+a)^2+18(x+a)+95-143+143k=0$? – Charlie Chang Jul 31 '20 at 00:20

5 Answers5

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Let's take one equation \begin{align*} 3x^2+7x+7 & \equiv 0 \pmod{11}\\ 4(3x^2+7x+7) & \equiv 4(0) \pmod{11}\\ x^2+28x+28 & \equiv 0 \pmod{11}&& (\because 4(3) \equiv 1 \pmod{11})\\ x^2+6x+6 & \equiv 0 \pmod{11}&& (\because 28 \equiv 6 \pmod{11})\\ (x+3)^2-3 & \equiv 0 \pmod{11}\\ (x+3)^2-5^2 & \equiv 0 \pmod{11}&& (\because 5^2 \equiv 3 \pmod{11})\\ (x-2)(x+8) & \equiv 0 \pmod{11} \end{align*} Since $11$ is prime so if $11 | ab$, then $11$ divides at least one of them, so we get $$x\equiv 2 \pmod{11} \quad \text{ or } \quad x \equiv -8 \equiv 3\pmod{11}.$$ Likewise (you can work this out yourself) $$3x^2+5x+4 \equiv 0 \pmod{13} \implies x\equiv 2 \pmod{13} \, \text{ or } \, x \equiv \color{blue}{b}\pmod{13}. $$ So we have the following situation \begin{align*} x&\equiv 2 \pmod{11} & x&\equiv 2 \pmod{11} & x&\equiv 3 \pmod{11} & x&\equiv 3 \pmod{11}\\ x&\equiv 2 \pmod{13} & x&\equiv \color{blue}{b} \pmod{13} & x&\equiv \color{blue}{b} \pmod{13} & x&\equiv 2 \pmod{13} \end{align*} Now use CRT (hopefully you know how to apply it to simple systems like these) to solve these systems.

For example the last system \begin{align*} x & \equiv 3 \pmod{11}\\ x & \equiv 2 \pmod{13} \end{align*} yields $$x \equiv 3(13)(6)+2(11)(6) \equiv \color{red}{80} \pmod{143}. $$ Likewise you will get a total of $\color{red}{4}$ incongruent solutions.

Anurag A
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  • Thank you very much, but $1=6\cdot 11-5\cdot 13$ so in the last part it shoud be: $$x \equiv 13\cdot 3\cdot (-5)+11\cdot 2\cdot 6 = -63 \equiv 80 \pmod {143}$$. – The student Jul 31 '20 at 07:33
  • @Thestudent What is wrote is correct because $6 \equiv -5 \pmod{11}$ and the inverse of $13$ modulo $11$ is $-5 \equiv 6$. – Anurag A Jul 31 '20 at 08:28
  • Yes I know it is right but I think it should be said, it is not so obvious $:)$. – The student Jul 31 '20 at 08:37
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hint

as $$95\equiv -48 \mod 143$$

the equation becomes $$3x^2+18x-48\equiv 0 \mod 143$$

but $$\delta=81+144=225=(15)^2$$ thus, it gives

$$3(x-2)(x+8)\equiv 0\mod 143$$

hamam_Abdallah
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HINT.-One has $3(x+3)^2+68\equiv0\pmod{11*13}$ so $$Y^2\equiv3\pmod{11}\\Y^2\equiv-1\pmod{13}$$ It follows $$Y\equiv5,6\pmod{11}\Rightarrow x\equiv2,3\pmod{11}\\Y\equiv5,8\pmod{13}\Rightarrow x\equiv2,5\pmod{13}$$ Thus $$x=2,57,80,135\pmod{143}$$ i.e. $x=2+143n$ for $n\in\mathbb Z$ and so is for $57,80$ and $135$.

Piquito
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You are allowed to use the quadraditic formula.

$3x^2 + 18x + 95 \equiv 0 \pmod {143}$ means

$x \equiv \frac {18\pm \sqrt{18^2 -4*95*3}}{6} \pmod {143}$.

Now use Chinese remainder theorem on $143 = 13*11$

$\frac {18\pm \sqrt {18^2 -4*95*3}}{6} \pmod {11}\equiv$

$\frac {7\pm {7^2 - 4*7*3}}{6}*12\pmod {11}\equiv$

$2(7\pm \sqrt{49 -12*7}) \equiv 14\pm 2\sqrt {5-7}\equiv$

$3 \pm 2\sqrt {-2}\equiv 3\pm 2\sqrt{9,64} \equiv 3\pm 6,16\equiv 9,8\pmod{11}$

And $\frac {18\pm \sqrt {18^2 -4*95*3}}{6} \pmod {13}\equiv$

$\frac {5\pm \sqrt {5^2 - 12*4}}{6}\pmod {13}\equiv$

$-12(\frac {5\pm \sqrt {25 + 4}}{6}\equiv -2(5\pm \sqrt{29})\equiv$

$-10 \pm2\sqrt{3}\equiv 3 \pm 2\sqrt{16,81}\equiv 3\pm 8,18\equiv 11,8$

So you have four solutions.

fleablood
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You can also perform all calculations in the $\text{modulo } 143$ system, but there is some algebra you have to use to verify you get all the solutions.

Using elementary number theory we find that $3^{-1} \equiv 48 \mod 143$, so multiplying through we can also, equivalently, examine

$\tag 1 x^2 + 6x + 127 \equiv 0 \mod 143$

and this can be written as

$\tag 2 (x + 3)^2 \equiv 25 \mod 143$

and we get an easy solution, $x + 3 \equiv 5 \mod 143$.

Using brute force logic we can write as true

$\quad x^2 \equiv 1 \mod 143 \iff$
$\quad\quad x \equiv 1 \mod 143 \,\lor\, x \equiv 12 \mod 143 \,\lor\, x \equiv 131 \mod 143 \,\lor\, x \equiv 142 \mod 143$

Multiplying out we get $4$ solutions to $u^2 \equiv 25 \mod 143$,

$\quad 5 \times 1 \equiv 5 \mod 143$
$\quad 5 \times 12 \equiv 60 \mod 143$
$\quad 5 \times 131 \equiv 83 \mod 143$
$\quad 5 \times 142 \equiv 138 \mod 143$

Thus,

$x=2,57,80,135 \mod 143$

CopyPasteIt
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