On $n! + k^2 = m^2$ Brocard's problem is the case $k=1$ of,
$$n!+k^2 = m^2\tag{1}$$
However, for general $k$,  
$$4!+1 = 5^2\tag{2}$$
$$6!+3^2 = 27^2\tag{3}$$
$$8!+9^2 = 201^2\tag{4}$$ 
and using the first two, we can observe two polynomial identities,
$$(x+1)(x+2)(x+3)(x+4) + 1 = (x^2+5x+5)^2\tag{5}$$
$$(2x+1)(2x+2)(3x+3)(2x+4)(2x+5)(3x+6)+3^2 = (12x^3+54x^2+72x+27)^2\tag{6}$$
$$(a_1x+1)(a_2x+2)\dots(a_8x+8)+k^2 \overset{\bf{?}}{=} P(x)^2\tag{7}$$
(The first is by Alain Verghote.) 
Question: Anybody knows how to find non-zero $a_i$ to $(7)$ using $(4)$, or using any of the 36 solutions to $8!+k^2=m^2$?
 A: Let's pretend I haven't seen any of the identities above. I just know the fact that $4!+1=5^2$ and I want to find an identity of this form:
$$
(ax+1)(bx+2)(cx+3)(dx+4) + 1 = P(x)^2
$$
Where $a, b, c, d \in \mathbb{N}$ and $P\in \mathbb{N}[x]$. Well, since $4!+1=5^2$, we know that the constant term of $P$ is $5$. So now let's rewrite the equation using polynomial $Q$, where $P(x) = Q(x) + 5$ (so the $Q$ doesn't have the constant term).
$$
(ax+1)(bx+2)(cx+3)(dx+4) + 1 = (Q(x)+5)^2
$$
$$
(ax+1)(bx+2)(cx+3)(dx+4)  = (Q(x)+4)(Q(x)+6)
$$
Now because of the prime factorization of $6$, the factor $Q(x)+6$ must be a product of $(bx+2)$ and $(cx+3)$ and so $Q(x)+4$ must be a product of $(ax+1)$ and $(dx+4)$.
$$
\begin{array}{rcl}
(ax+1)(dx+4) &=& adx^2 + (4a+d)x+4 \\
(bx+2)(cx+3) &=& bcx^2 + (3b+2c)x+6 \\
\end{array}
$$
Notice that $Q(x)+4$ and $Q(x)+6$ differ only in the constant term so the following equations must hold:
$$
\begin{array}{rcl}
ad &=& bc \\
4a+d &=& 3b+2c \\
\end{array}
$$
Now, every such solution $(a, b, c, d)$ should form a suitable polynomial. For example $(a, b, c, d) = (7, 7, 7, 7)$ is a solution. For this quadruple we get the equality:
$$
(7x+1)(7x+2)(7x+3)(7x+4) + 1 = (49x^2 + 35x + 5)^2
$$
Quite amusing but I think it can be aplicable to the higher degrees. Again, for the fact $8!+9^2=201^2$ we use the same technique and the equation will now look like this:
$$
(ax+1)(bx+2)(cx+3)(dx+4)(ex+5)(fx+6)(gx+7)(hx+8) + 9^2 = (Q(x)+201)^2
$$
$$
(ax+1)(bx+2)(cx+3)(dx+4)(ex+5)(fx+6)(gx+7)(hx+8) = (Q(x)+192)(Q(x)+210)
$$
From the prime factorisation of $192$ and $210$ we see that $Q(x)+210$ can be only equal either to $(ax+1)(ex+5)(fx+6)(gx+7)$ or $(bx+2)(cx+3)(ex+5)(gx+7)$. And now we would just need to create that set of (four) equations and find any solution $(a, b, c, d, e, f, g, h)$.
But that was too much pain for me... :-/
