I'm unsure if this question is even interesting. I am playing around with partition numbers $p(n) :=$ # partitions of $n$, and I noticed that $p(n)$ never really is a square number, except for of course $n = 0, 1$. Is there a simple reason for this?
It also seems to be similar for other modified partition numbers, such as $p_{t}(n) :=$ #t-regular partitions of $n$ (none of its parts are divisible by $t$), at least for $t = 2, 3, 4$ and a few more. In these cases the only square $p_{t}(n)$ are when $p_{t}(n) \in \{0,1\}$. For example, $p_{11}(n)$ is square (apparently) only when $0 \leq n \leq 6$, in which case $p_{11}(n) \in \{0, 1\}$.
It gets a little weirder when we consider $p_{s, t}(n) :=$ #s-regular and t-distinct partitions of $n$ (no part appears $t$ or more times). For example, when $(s, t) = (7, 9)$ we have $p_{7, 9}(0) = 1 = p_{7, 9}(1)$ and $p_{7, 9}(11) = 49$, $p_{7, 9}(44) = 35721 = 189^2$. (It is also somewhat interesting that the input/outputs are either divisible by 7 or 9). These are the only ones up to $10^4$ (very slow program so hard to go past this). For convenience, the generating function for $p_{7, 9}(n)$ is $\prod_{n\geq 1} \frac{(1-q^{7n})(1-q^{9n})}{(1-q^{n})(1-q^{63n})}$. Somewhat similar results for other pairs of $(s, t)$.
Maybe this has to do with the generating functions?