Determining whether this congruence is solvable: $3x^2+6x+5 \equiv 0\pmod{89}$ I'm trying to figure out whether the following quadratic congruence is solvable: $3x^2+6x+5 \equiv 0\pmod{89}$.
It's impossible to divide $3x^2+6x+5$ to a form of $f(x) \cdot g(x)=3x^2+6x+5$ and then to check whether $f(x)\equiv 0
\pmod{89}$ or $g(x)\equiv 0(89)$, but $3x^2+6x+5 \equiv 0\pmod{89}$ is equal to $3(x+1)^2+2 \equiv 0\pmod{89}$ or $3(x+1)^2 \equiv -2\pmod{89}$ or $3(x+1)^2 \equiv 87\pmod{89}$ or
$(x+1)^2 \equiv 29\pmod{89}$. for $y=x+1$, I need to determine whether  $y^2 \equiv 29\pmod{89}$ is solvable, and it is not. Am I able to conclude something about the original equation in this way? what is the correct way to solve this problem?
Thanks a lot!
 A: Yes, your inference is correct. Essentially it is a special case of the well-known discriminant test. Namely, if a quadratic $\rm\:f(x)\in R[x]\:$ has a root in a ring R, then its discriminant is a square in R. Said contrapositively, if the discriminant is not a square in R, then the quadratic has no root in R.
The proof by completing the square works in any ring R (so in $ \mathbb Z/89 = $ integers mod $89$), viz.
$$\rm\: \ \ 4a\:(a\:x^2 + b\:x + c = 0)\:\Rightarrow\: (2a\:x+b)^2 =\: b^2 - 4ac $$
When learning about (modular) arithmetic in new rings it is essential to keep in mind that, like above, any proofs from familiar concrete rings (e.g. $\mathbb Q,\mathbb R,\mathbb C)$ will generalize to every ring if they are purely ring theoretic, i.e. if the proof uses only universal ring properties, i.e. laws that hold true in every ring, e.g. commutative, associative, distributive laws. Thus many familar identities  (e.g. Binomial Theorem, difference of squares factorization) are universal, i.e. hold true in every ring.  
This is one of the great benefits provided by axiomatization: abstracting the common properties of familiar number systems into the abstract notion of a ring allows one to give universal proofs of ring theorems. It is not necessary to reprove these common ring properties every time one studies a new ring (such reproofs occurred frequently before rings was axiomatized).
A: If you know about the law of quadratic reciprocity, it gives you a way to tell whether $y^2\equiv29\pmod{89}$ is solvable. 
Here's how it goes: Since 29 and 89 are both prime and congruent 1 mod 4, $y^2\equiv29\pmod{89}$ is solvable if and only if $y^2\equiv89\pmod{29}$ is solvable. That reduces to $y^2\equiv2\pmod{29}$. Then there's a little result that says if $p$ is an odd prime then $y^2\equiv2\pmod p$ is solvable if and only if $p\equiv\pm1\pmod8$. 
There's more to quadratic reciprocity than what I've done here. It's in every intro number theory text, and all over the web. 
