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).
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 $q$.