Solve $x^2+3y = u^2$ and $y^2+3x=v^2$ in positive integers.

Question

The question is from the pg - 59 from ' An Introduction to Diophantine Equations ' by Titu Andreescu , Dorin Andrica , Ion Cucurezeanu.

Example 1 : Solve in positive system of equations in positive integers
$$\begin{cases} x^2+3y = u^2 \\ y^2 + 3x = v^2 \end{cases}$$
$\;\;\;\;\;\;\;\;\;\,\,\,\,\,\,\,\,\,\,\,\,\text{(Titu Andreescu)}$

Solution. The inequality $x^2 + 3y ≥ (x + 2)^2 , y^2 + 3x ≥ (y + 2)^2$ cannot both be true, because adding them would yield a contradiction.

So at least one of the inequalities $x^2 + 3y < (x + 2)^2$ and $y^2 + 3x < (y + 2)^2$ is true. Without loss of generality, assume that $x^2 + 3y < (x + 2)^2$.

Then $$x^2 < x^2 + 3y < (x + 2)^2 \implies x^2 + 3y = (x+1)^2$$ or, $3y = 2x+ 1$ . We obtain $x = 3k + 1, y = 2k + 1$ for some nonnegative integer $k$ and $y^2 + 3x = 4k^2 + 13k + 4$.

For $k > 5, (2k+ 3)^2 < 4k^2 + 13k+ 4 < (2k+ 4)^2$ ; hence $y^2 + 3x$ cannot be a perfect square. Thus we need only consider $k ∈ {\{0, 1, 2, 3, 4\}}$ . Only $k = 0$ makes $y^2 + 3x$ a perfect square; hence the unique solution is $$x = y = 1,\;\;\;\;\;\; u = v = 2.$$

But if we take , $$4k^2+13k + 4 = v^2$$ $$\implies k = \dfrac{-13 \pm\sqrt{105+16v^2}}{8}$$

Since $105+16v^2 = a^2 \implies 105 = (a-4v)(a+4v)$ which gives $a \in \{\pm11 , \pm13 , \pm19 ,\pm53\}$ . Out of these , only $a \in \{ \pm13 , \pm53\}$ works which gives $k=0,5$ , And so the the answer should be $$(x,y,u,v) = (1,1,2,2)\;\;\;,(16,11,17,13)\;\;\;\;,(11,16,13,17)$$

Who is correct here?

Answer 1

$\begin{cases} x^2+3y = u^2 \\ y^2 + 3x = v^2 \end{cases}$ -----(1)

"OP" gave numerical solution to equation (1) as:

$(v,u,x,y)=(2,2,1,1)=(13,17,16,11)$

There is another numerical solution & is given below:

$(v,u,x,y)=[(10),(25/4),(13/4),(19/2)]$

Answer 2

You are indeed correct; it is not hard to verify that your solution(s) are indeed solutions. The proof that you quote overlooks the case $k=5$, which yields your two additional solutions.