# $x^2 + 4x - \lambda +2 \equiv 0 \pmod 7$

I solved this problem but I don't know if my solution is right:

Find the values of $$\lambda, 0 \le \lambda \le 6$$, such that the congruence $$x^2 + 4x - \lambda +2 \equiv 0 \pmod 7$$ has a solution. Find all the solutions for the minimum value of $$\lambda$$ for which the congruence is solvable.

I have to solve the system of congruences $$\cases{y^2 \equiv b^2-4ac \pmod p\\ 2ax \equiv y-b \pmod p}$$

$$y^2 \equiv 4^2-4(2-\lambda) \equiv 1+4\lambda \pmod 7$$, so I need to know when $$1+4\lambda$$ is a quadratic residue $$\pmod 7$$.

For the Euler's criterion I know that $$(1+4\lambda |7) \equiv (1+4\lambda)^3 \equiv 1+64 \lambda^3 +12\lambda+12\lambda^2 \equiv 1+ \lambda(\lambda^2+5\lambda+5) \pmod 7$$.

But $$5$$ is not a quadratic residue $$\pmod 7$$ because $$(5|7) = (7|5)(-1)^6=(2|5)=(-1)^3=-1$$, so the quadratic equation $$\lambda^2+5\lambda+5 \equiv 0 \pmod 7$$ is not solvable.

I conclude that the unique value for which $$x^2 + 4x - \lambda +2 \equiv 0 \pmod 7$$ has a solution is $$\lambda =0$$.

My solutions are therefore $$\cases{y^2 \equiv 1 \pmod 7 \longrightarrow y \equiv \pm 1 \pmod 7 \\ 2x \equiv -5 \pmod 7 \longrightarrow x \equiv 1 \pmod 7\\ 2x \equiv -3 \pmod 7 \longrightarrow x \equiv 2 \pmod 7}$$

Is my solution right? Thanks you for all your suggestions.

No! There are four values of $$\lambda$$ such that $$x^2+4x-\lambda+2\equiv 0\pmod 7$$ has a solution $$x\in\mathbb{Z}$$.
Note that $$x^2+4x-\lambda+2=(x+2)^2-(\lambda+2)$$, so you want $$\lambda+2$$ to be a square, i.e., $$\lambda+2\equiv 0,1,2,4\pmod{7}$$.