$a^2+nb$ and $b^2+na$ are both squares

Question

1. Prove or disprove there is a positive integer $n$ such that there are infinitely many pairs $(a,b)$ of positive integers such that $a^2+nb$ and $b^2+na$ are perfect squares

I have made another question. Please try them.

Answer 1

For any given $n$, there are only finitely many solutions for $a,b$. We have: $nb=(x-a)(x+a)$ and $na=(y-b)(y+b)$.

Let $n=p_1p_2=q_1q_2$ and $b=r_1r_2,a=s_1s_2$ such that $x-a=p_1r_1,x+a=p_2r_2$, $y-b=q_1s_1,y+b=q_2s_2$.

We get: $p_2r_2-p_1r_1=2s_1s_2$ and $q_2s_2-q_1s_1=2r_1r_2$.

In particular, $p_2r_2>2s_1s_2$ and $q_2s_2>2r_1r_2$; multiplying the two inequalities gives: $4r_1s_1<p_2q_2$.

For fixed $n$, the number of choices of $p_2,q_2$ is finite and therefore the number of choices of $r_1,s_1$ is also finite. For a fixed choice of these values, we can solve uniquely for $a,b,x,y$ from the four linear equations we have for these variables (or for $r_2,s_2$ since the equations for these two variables can be easily seen to be not proportional).

Answer 2

$a^2+nb = u^2\tag{1}$
$b^2+na = v^2\tag{2}$

From equation $(1)$ and $(2)$, solving the simultaneous equations $$\begin{cases} u+a=nt\\ u-a=b/t\\ v+b=ns\\ v-b=a/s\\ \end{cases} $$

then we get a solution as follows.
There is a positive integer solution for $n = -1+4st$.

$n = -1+4st$
$a = -s(-2t^2+s)$
$b = t(-t+2s^2)$
$u = -t+2st^2+s^2$
$v = t^2+2ts^2-s$
s,t are arbitrary.

Example:

                     [n,a,b,u,v]
                     
                     [3, 1, 1, 2, 2]
                     [15, 12, 12, 18, 18]
                     [23, 32, 15, 37, 31]
                     [31, 60, 16, 64, 46]
                     [39, 96, 15, 99, 63]
                     [23, 15, 32, 31, 37]
                     [35, 45, 45, 60, 60]
                     [47, 87, 56, 101, 85]
                     [59, 141, 65, 154, 112]
                     [31, 16, 60, 46, 64]
                     [47, 56, 87, 85, 101]
                     [63, 112, 112, 140, 140]
                     [79, 184, 135, 211, 181]
                     [39, 15, 96, 63, 99]
                     [59, 65, 141, 112, 154]
                     [79, 135, 184, 181, 211]
                     [99, 225, 225, 270, 270]