Finding an isomorphism between polyomial quotient rings Let $F_1  = \mathbb{Z}_5[x]/(x^2+x+1)$ and $F_2 = \mathbb{Z}_5[x]/(x^2+3)$. Note neither $x^2+x+1$ nor $x^2+3$ has a root in $\mathbb{Z}_5$, so that each of the above are fields of order 25, and hence they are isomorphic from elementary vector space theory, however I'm tasked with constructing an explicit isomorphism between the two fields, so I applied the technique I've learned here and elsewhere as follows.
First, let $\phi: F_1 \rightarrow F_2$ by $x \mapsto ax+b$ (as the base field is fixed it suffices to find the image of $x$). Then we should have that $\phi(x^2+x+1) = x^2 + 3$, and the left hand side reduces to $\phi(x)^2 + \phi(x) + 1$ as $\phi$ is a homomorphism. Simplifying, we have $(ax+b)^2 + (ax+b) + 1 = a^2x^2 + 2abx + b^2 + ax+b+1 = a^2x^2 + (2ab+a)x + (b^2+b+1) = x^2+3$. 
However, this appears to be a problem. Notably, since $a^2 = 1$ mod 5, we have $a = 1,4$. Picking the former, we have $2b+1 = 0$ and thus $b = 2$, a problem as $2^2+2+1 = 7 \neq 1$ mod 5. Trying $a=4$, we have $3b+4 = 0$ so $b = 2$, and again the same problem. 
Are there flaws in my technique or arithmetic that I don't see? If not, what other options do I have to attempt to construct the isomorphism?
 A: HINT:
You want $(\mathbb{Z}/5)[x]/(x^2+x+1) \to (\mathbb{Z}/5)[x]/(x^2+3)$,  $x \mapsto a x + b$, so  $x^2+x+1 \mapsto (ax+b)^2+ (ax + b) +1$, and you want the latter a  multiple of $x^2+3$. Divide $(ax+b)^2+ (ax + b) +1$ by $x^2+3$ and you get the remainder 
$$a\,(2 b+1)\, x + (b^2 + b+1 - 3 a^2)$$
A: As per Jyrki Lahtonen's hint, I will instead try $\phi(x^2+x+1) = c(x^2+3)$. So then since $x \mapsto ax+b$, we have $a^2x^2 + (2ab+a)x + b^2+b+1 = cx^2 + 3c$, so $a^2 = c$, $2ab+a = 0$, and $b^2+b+1 = 3c$. Since $a^2 = c$ and the only squares mod 5 are 1 and 4, we have that $c = 1,4$. Having already gleaned that 1 will not suffice, let us try $c = 4$.
Then $a^2 = 4$ so $a = 2,3$. The former did not yield nice results, so let $a = 3$. Then we have that $b + 3 = 0$ so that $b = 2$, and from the last relation $7 = 12$ which is true mod 5.
Therefore $a = 3$ and $b = 2$, so $x \mapsto 3x+2$ is an isomorphism.
A: Hint $\ $ Complete the square $\ 4(x^2\!+\!x\!+\!1) = (2x\!+\!1)^2\!+\!3 = X^2\!+\!3,\ $ for $\,X = 2x\!+\!1$
