When is $20q^4-40q^3+30q^2-10q$ a square for positive integer $q$? 
For what $q$ is the following polynomial a square?
  $$
\begin{align}
&20q^4-40q^3+30q^2-10q\\
=\:&10q(q - 1)(2q^2 - 2q + 1)
&q\in\mathbb N
\end{align}
$$

I know of two single cases, $q=1$ gives $0$ and $q = 2$ gives $100$. I tested $3\le q\le12$ and found none.
This is what I've tried so far:
I found that
$$
\begin{align}
\gcd(q,q - 1) &= 1\\
\gcd(q, 2q^2 - 2q + 1) &= 1\\
\gcd(q - 1, 2q^2 - 2q + 1) &= 1
\end{align}
$$
The problem is the same as determining whether there exists a $t\in\mathbb N$ such that
$$
q(q - 1)(2q^2 - 2q + 1) = 10t^2
$$
If $q$ is even, write $q=2p$ and say
$$
p(2p-1)(8p^2-4p+1) = 5t^2
$$
The residues modulo $5$ are as follows
$$
\begin{align}
p\equiv0\pmod{5}\implies p(2p-1)(8p^2-4p+1)\equiv0\pmod{5}\\
p\equiv1\pmod{5}\implies p(2p-1)(8p^2-4p+1)\equiv0\pmod{5}\\
p\equiv2\pmod{5}\implies p(2p-1)(8p^2-4p+1)\equiv0\pmod{5}\\
p\equiv3\pmod{5}\implies p(2p-1)(8p^2-4p+1)\equiv0\pmod{5}\\
p\equiv4\pmod{5}\implies p(2p-1)(8p^2-4p+1)\equiv4\pmod{5}
\end{align}
$$
Which means that $p\not\equiv4\pmod5$, and we now need to test if it's a square in the rest of the cases.

Case r = 5p
In this case we are solving $r\cdot(10r - 1)(200r^2 - 20r + 1) = t^2$ and since the factors are coprime, they must all be squares, however modulo $7$ atleast one factor is not one of the quadratic residues for all cases.

Case r = 5p + 1
In this case we are solving $(5r+1)(10r+1)(40r^2+12 r+1) = t^2$ and since the factors are coprime, they must all be squares, however modulo $3$ atleast one factor is not one of the quadratic residues for cases $1$ and $2$, and the last case $0$ we have an observed solution at $q=2$, however for $q=12$ we don't get a square, so it remains to prove whether there are other than $r=0$ which are square.

At this point I stopped because there are many cases, and I don't know how to prove that it is only for $r=0$ when $r=5p+1$ that the polynomial is square, or if there are other such $r$.
So the thing I'm asking specifically is: Is there a more elegant way, and if not, how do I prove that the observed cases are the only solutions (or if not, what is the set of solutions).
 A: Let $r=q(q-1)/2$, then $8r^2+2r=10t^2$, $(8r+1)^2=80t^2+1$.  So you might solve the Pellian equation $k^2=80t^2+1$, and then try see if any of the $k$ have the right form in terms of $q$.
A: A similar Pell approach. 
We want solutions to 
$$10q(q-1)(2q^2-2q+1)=n^2$$
So $q\in{0,1}$ with $n=0$ are trivial examples. We might get solutions as linear combinations of these. 
Let $2p=q(q-1)$.
Then $$5(4p)(4p+1)=n^2$$
We know $n$ must be a multiple of 5, so let n=5k.
$$5(4p)(4p+1)=25k^2$$
$$(4p)^2+4p=5k^2$$
Let x=4p and Complete the square:
$$(x)^2+x+\frac{1}{4}=5k^2+\frac{1}{4}$$
$$ (x+\frac{1}{2})^2-5k^2=\frac{1}{4} $$
Then multiply by 5: 
$$(2x+1)^2-5(2k)^2=1$$
Since $x=4p$, and $2p=q(q-1)$,
$$[4q(q-1)+1]^2-5(2k)^2=1$$
With factoring:
$$ (2q-1)^4-5(2k)^2=1$$
We get a Pell equation: $m^2-5n^2=1$
Initial values for $(m,n)=(9,4)$ implying $q=2$ and $k=2$, so $n=10$
Then you can apply the Brahmagupta Solution to Pell's equation which allows one to generate infinitely many solutions from a single one. 
Given $x^2-py^2=1$ and solution $(x_0,y_0)$:
$$x_{n+1}=x_0x_n+py_0y_n$$
$$y_{n+1}=y_0x_n+x_0y_n$$
So you want $x_n$ that are odd perfect squares. 
