When $108n^4+1$ is a perfect square if $n$ is an integer? I want to find which integers values of $n$ makes $108n^4+1$ a perfect square
One solution is $n=0$, and I already proved that $n$ is not odd, I believe that the only solution is $n=0$, is there a counterexample?
If $n$ is odd, $108n^4+1$ would be congruent $5$ modulo 8 and that is impossible for it to be a perfect square.
 A: This is a long-winded comment.  
This is not an answer.
First of all, I am repeating the comments that I left, after the posted question.
The primitive solution to $X^2 - 3Y^2 = 1 $ is $(X,Y) = (2,1).$ All solutions $(X,Y)$ are generated by the formula
$$\left(X_k + Y_k\sqrt{3}\right) = \left(2 + \sqrt{3}\right)^k ~: k \in \Bbb{Z^+}. \tag1 $$
As a necessary and sufficient condition of the problem, you are looking for a value $(X_k,Y_k)$, generated by (1) above, where $Y_k$ has form $(6n^2).$  This is because $(108n^4)$ can be re-expressed as $3 \times \left( ~6 ~\left[n\right]^2 \right)^2.$
The next element in the sequence, generated by (1) above is $(X_2,Y_2) = (7,4).$  Note that this element is of form $(A,B)$, where $A$ is odd and $B$ is even.
If you consider any such expression
$$\left(A + B\sqrt{3}\right) \times \left(2 + \sqrt{3}\right) ~: ~A ~\text{odd}, ~B ~\text{even},$$
you will see that the result is
$$\left(2A + 3B\right) + \sqrt{3}\left(2B + A\right),$$
where $(2A + 3B)$ is even and $(2B + A)$ is odd.
Then, if you take any expression of the form
$$\left(R + S\sqrt{3}\right) \times \left(2 + \sqrt{3}\right) ~: ~R ~\text{even}, ~S ~\text{odd},$$
you will see that the result is
$$\left(2R + 3S\right) + \sqrt{3}\left(2S + R\right),$$
where $(2R + 3S)$ is odd and $(2S + R)$ is even.

This means that if you examine the sequence formed by (1) above, the elements in that sequence will have form
$$\left(\text{even}, ~\text{odd}\right), ~\left(\text{odd}, ~\text{even}\right), ~\left(\text{even}, ~\text{odd}\right), ~\left(\text{odd}, ~\text{even}\right), \cdots .$$
Since the 2nd component $Y_k$ is required to be of form $6n^2$, it is required to be even.  Therefore, when examining the sequence of elements generated by (1) above, all such elements generated when $k$ is an odd positive integer may automatically be rejected.
Therefore, you are looking for some elements $R_k,S_k$ generated by
$$\left(R_k + S_k\sqrt{3}\right) = \left(2 + \sqrt{3}\right)^{(2k)} = \left(7 + 4\sqrt{3}\right)^k. \tag2 $$
The first three elements of the sequence generated by (2) above are
$$(7,4), ~(97,56), ~(1351,780).$$

At this point, in desperation, my analysis goes in the following direction:
The $S_k$ component generated by
$$\left(R_k + S_k\sqrt{3}\right) = \left(7 + 4\sqrt{3}\right)^k$$
may be computed as
$$\sum_{i \in \Bbb{Z^+}, ~i ~\text{odd}, ~i \leq k} \left[ ~\binom{k}{i} \times 7^{(k - i)} \times 4^i \times 3^{[(i-1)/2]} ~\right]. \tag3 $$
Examining the terms expressed in (3) above, you see that whenever $i \geq 3$, then the term is a multiple of $(3)$.  Further, the first term,
$$\binom{k}{1} \times 7^{(k-1)} \times 4^1$$
will be a multiple of $(3)$ if and only if $k$ is a multiple of $(3)$.
Therefore, attention may be restricted to the elements generated by the formula in (2) above, with $k$ required to be a multiple of $(3)$.
This implies that you may restrict your attention to elements generated by
$$\left(U_k + W_k\sqrt{3}\right) = \left(1351 + 780\sqrt{3}\right)^k. \tag4 $$
(4) above is as far as my thinking takes me.  I see no simple way of demonstrating that none of the elements generated by (4) above will have a 2nd component $W_k$ of form $6(n^2).$
