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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.

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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., $\lambda+2\equiv 0,1,2,4\pmod{7}$.

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