What do these extra solutions mean? I'm trying to find a constant $a$ such that $n(n+1)(n+2)(n+3)$ is equivalent to $(n^2+an)(n^2+an+2)$.
Clearly by inspection, we have $a=3$. However, say I wish to substitute $n=-1$. We get our desired solution, but we also have another value $a=-1$. 
My question is, what does this extra value mean, and why does it appear there?
 A: $a = -1$ satisfies the equation only if $n = -1$. It does not hold true for all values of $n$. On the other hand, $a = 3$ satisfies the equation for all values of $n$.
You can write an equation which expresses $a$ in terms of $n$ (I do not know how hard this will be). This equation will allow you to determine the values of $a$ which will "work" for each corresponding value of $n$.
However the solutions of $a$ for a particular value of $n$ may not work for all other values of $n$. Of course, $a = 3$ will always be a result inferable from the equation derived.
A: Solving $$n(n+1)(n+2)(n+3)=(n^2+an)(n^2+an+2)$$
for a yields
$$a=-2n-\frac{2}{n}-3$$
when $n\neq0$
A: Subtracting the first expression from the second and factoring gives $n(a-3)(na+2n^2+3n+2) = 0$.  Therefore, there are three solutions we should consider: 
\begin{align*}
n &= 0\\
a &= 3\qquad\text{and} \\
a &= -(2n+3+2/n) \leftrightarrow n=\frac{-(3+a)\pm\sqrt{(a-1)(a+7)}}4
\end{align*}
Which of these three you want would depend on the precise wording of the problem.  If the problem asked for "all $a$ and $n$", you’d want all three, and a graph in the $an$-plane may be a nice addition to the solution (by the way, $n=-1$ gives $a=1$).  If the problem asked for "$a$ such that ... for all $n$", you’d want the second.  If the problem asked for "$n$ such that ... for all $a$", you’d want the first.  In this context, “is equivalent to” means “equals … for all [other variables]”, so the problem could be written “find $a$ such that … equals … for all $n$”, and the answer is $a=3$.
