<s>[INST] Product of $i^4+4$ is a perfect square The following is the Problem $50$ (Folklore) from this pdf:

Determine all positive integers $k$ such that $\prod_{i=1}^k (i^4+4)$ is a perfect square.

I realised that $i^4+4$ is just the Sophie-Germain identity, so the problem resolves to proving $2(k^2+2k+2)(k^2+1)$ is square, and I realised $(k^2+2k+2)(k^2+1)=(k^2+k+1)^2+1,$ so we want to prove $(k^2+k+1)^2+1=2n^2$ (because if $2x$ is a perfect square, $x$ is $n^2$). This makes me think of the Pell equation, but it seems to be too complicated.
Any help would be much appreciated, thanks in advance.
 [/INST] I suppose this question requires to find all integers $k$ satisfying the statement than proving that statement is true ($k=1$ is false since it evaluates to 5)
So we can start from $2(k^2+1)(k^2+2k+2)$ which you found. For perfect squares we need two factors of the number to be equal:
$2(k^2+1)=k^2+2k+2_{---(1)}\\OR\\2(k^2+2k+2)=k^2+1_{---(2)}\\OR\\(k^2+1)(k^2+2k+2)=2_{---(3)}$
Solving $(1)$, we get $k=2 \: or \: 0$
Solving $(2)$, we get $k=-3 \: or \: -1$
Solving $(3)$, we get $k=0 \: or \: -1$
We can reject the solutions 0, -1 and -3 since $k\geqslant1$, therefore the only answer is 2.
Edit: Still figuring out how to find the integers k for non-equal factors like $2\times8=4^2$
For others who don't know how to find $2(k^2+1)(k^2+2k+2)$
$i^4 + 4\\= i^4 + 4i^2 + 4 - 4i^2\\=(i^2+2)^2-(2i)^2\\=(i^2+2i+2)(i^2-2i+2)\\=[(i+1)^2+1][(i-1)^2+1]$
So the question resolves to finding integers k making $\prod_{i=1}^{k}[(i+1)^2+1][(i-1)^2+1]$ a perfect square
Observe that for the graphs $L_1:y=(x+1)^2+1$ and $L_2:y=(x-1)^2+1$, $L_1$ is just 2 units on the left of $L_2$ which means substituting some n+2 into $L_2$ gives the same result as substituting n into $L_1$
Then we can get back to the given expression:
$\prod_{i=1}^{k}(i^4+4)\\=\prod_{i=1}^{k}[(i+1)^2+1][(i-1)^2+1]$
For $k\geqslant3$, there will be $k-2$ squares in the middle of the expansion. So the expressions unknown to be perfect squares are the two terms of both ends:
i.e. $[(1-1)^2+1][(2-1)^2+1] \\ [(k-1+1)^2+1][(k+1)^2+1]\\=2(k^2+1)(k^2+2k+2)$
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