Determine $x$ such that $2x^2 - x - 6$ be a multiple of $53$

Question

I used the modular arithmetic : $$2x^2 - x - 6 \equiv 0 \mod 53$$ and then : $$(2x+3)(x-2) \equiv 0 \mod 53$$

I'm stuck here and don't know what to do with $AB \equiv 0 \mod C$

Answer 1

Since $53$ is prime, it divides either $x-2$ or $2x+3$. Thus $x\equiv 2,\,25\,(\operatorname{mod}53)$.

Answer 2

It follows from your computations that $x=2$ will work.

If you're after all solutions (modulo $53$), you do\begin{align}(2x+3)(x-2)\equiv0\pmod{53}&\iff 2x+3\equiv0\pmod{53}\vee x-2\equiv0\pmod{53}\\&\iff2x\equiv50\pmod{53}\vee x\equiv2\pmod{53}\\&\iff x\equiv25\pmod{53}\vee x\equiv2\pmod{53}.\end{align}

Answer 3

See Euclid's Lemma. $53$ divides the product $(2x+3)(x-2)$. Then $53$ divides one of them, or both. You now have all the cases.

Answer 4

$53=1*53$(it is a prime number)

So once let, $2x+3=53$
And second time let $x-2=53$
For (i) $x=25$
$\Rightarrow$ $Number=53*27$

Similarly for (ii) $x=55$
$\Rightarrow$ $Number=53*113$

Now considering another case when $53|2x+3$ or $53|x-2$
Which means (i) $53n=2x+3$ (Here n can take any odd integral value)
OR
(ii) $53n=x-2$(here n can take any integral value)

It means there can't be any discrete value/s of x

Answer 5

You can do also like this. Multiply this with $8$ $$2x^2 - x - 6 \equiv 0 \mod 53$$ so $$16x^2 - 8x - 48 \equiv 0 \mod 53$$ so $$16x^2 - 8x +1- 49 \equiv 0 \mod 53$$ so $$(4x-1)^2 \equiv 7^2 \mod 53$$ so $$4x-1 \equiv \pm 7 \mod 53$$ ...

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Note that if you have $$ax^2+bx+c\equiv 0 \mod d$$ and $\gcd(2a,d) =1$ then you can do this always (multiply with $4a$): $$4a^2x^2+4abx+4ac\equiv 0 \mod d$$ so $$4a^2x^2+4abx+b^2\equiv b^2-4ac \mod d$$ $$(2ax+b)^2\equiv \Delta \mod d$$ where $\Delta = b^2-4ac$.