i think for all n=3^k ?

Factorise 2x^2+5xy+2y^2 into (2x+y)(x+2y) We can write 2x+y = a and x+2y = n/a where a|n Do three cases, which are n=0, 1, or 2 (mod 3) For n=0 mod 3, we will get the solutions n=9k, For n=1 mod 3, there will be no solutions n, For n=2 mod 3, we will get that every n in this case are solutions. Hence, we get n=9k and n=3k+2 for all integers k

solutions

$2x^2 + 5xy + 2y^2 = n$ $(x+y)^2+xy=n-(x+y)^2$ by comparing $(x+y)^2$ with $-(x+y)^2$ we get $n=-k^2$ then integer solutions are $x=k$ and $y=-k$ or vise versa