How Do I solve the below congruence?

Question

let $x$ be an integer, I want to solve the following quadratic equation satisfying the below condition :

$ 5x^2+40x+100=4 \mod m$ , with $m$ is integer , such that $m=6,7,8,9$

and $ \; 5x^2+40x+100=0 \mod m$ with $m =2,4,5$

and $\; 5x^2+40x+100=1\mod m$, with $m=3$ .

I have tried the Chinese remainder theorem but I didn't succeed, Any way?

Note:I have edited the equation because i have got a wrong typo without changing the meaning of question

Answer 1

Hint:

Solve equation(s) modulo primes first, i.e. modulo $2, 3, 5, 7$, check whether the discriminant is a square.

Modulo $6$, from a solution $\alpha\bmod 2$ and a solution $\beta\bmod 3$, you can deduce a solution $\bmod 6$ using a Bézout's identity between $2$ and $3$ (e.g. $3-2=1$): $$\xi =3\alpha-2\beta\bmod 6.$$ For solutions mod. $2^3$ and $3^2$, you deduce them from the solutions $\bmod 2$ and $3$ tespectivelu=y using Hensel's lifting.

Answer 2

(1) For $m=6,$ write the equation as $6x^2-x^2+40x-96=0,$ which becomes $x^2-4x-=0$ modulo $6.$ This factors to give $x(x-4)=0,$ from which you may proceed as usual. For $m=7,8,9,$ similar maneuvers work.

(2) This is particularly easy, as it reduces respectively to $x^2=0,\,x^2=0$ and $0=0$ respectively modulo $2,4,5.$

(3) This reduces to $2x^2+x=0,$ which also factors nicely, etc.

Answer 3

In my Humbel Guess :let $y=5x^2+40x+100=(2x+10)^2+x^2$ , The Nice particular solution for that equation satisfying the titled conditions is $y=2020$ and $2020$ can be written as $2020=16^2+42^2$ , now $42=2x+10$ implies $x=16$