Yesterday, i discovered a nice thing while playing with numbers. It is trivial to note that $\forall n\in \mathbb{Z^+},\exists x,y\in \mathbb{Z}$ such that $3^n=5x^2+y$ has solutions. Also, undoubtedly, $x,y$ are of different parity. But the good thing i noticed is really tough to prove i think.
Conjecture
If we choose to consider only the $x,y$ such that $\mid x-y \mid$ is minimum, then if we make a table whose rows are the different values of this predefined $x,y$ for consecutive values of $n$, like this
where we make another row $y_{i-1}+y_i$(mainly for convenience) which will attain the sum of the current $y ~+$ the $y$ of the previous entry. Now, interestingly, if we fix $y_i$, then i noticed upto $n=11$ using my calculator that $y_{i-1}+y_i\equiv y_{i+1}[5]$
(I purposely did not add the $n=11$ because i love even numbers :D But for $n=11$, $(x,y)=(188,427)$ and again $-518 \equiv 427 [5]$) I cannot understand the reason of its occurence. Can there be any elementary proof? Thanks for giving time!
where the two ends of $/$ denotes congruency between the two $\mod 5$
I think minimality of $|x-y|$ doesn't matter here. Let $(x_{i-1}, y_{i-1}), (x_{i}, y_{i}), (x_{i+1}, y_{i+1})$ be any solutions of $3^{n} = 5x^{2} + y$ for $n = i-1, i, i+1$, respectively. Then $3^{n} \equiv y_{n}\,(\mathrm{mod} \,5)$, so $$ y_{i+1} - (y_{i} + y_{i-1}) \equiv 3^{i+1} - 3^{i} - 3^{i-1} = 3^{i-1}(9 - 3 - 1) = 5 \cdot 3^{i-1} \equiv 0 \,(\mathrm{mod}\,5). $$