Find all natural numbers $x$ and $y$ satisfying the equality $19(x^2+y^2)=221+33xy$

Question

I think the substitution $x=\xi+\eta,$ $y=\xi-\eta$ can be done. Then the equation takes the form $$ \begin{gathered} 38(\xi^{2}+\eta^{2})=221+33(\xi^{2}-\eta^{2}) \\ 5 \xi^{2}+71 \eta^{2}=221 \end{gathered} $$

whence $\xi^{2}=30-71 n$, $\eta^{2}=1+5 n$. For $n=0$ we obtain noninteger solutions and for the rest one of the equalities has a negative right-hand side. Am I wrong?

Answer 1

We have $$19\left(x-\frac{33y}{38}\right)^2+\left(19-19\cdot\left(\frac{33}{38}\right)^2\right)y^2=221,$$ which gives $$\left(19-19\cdot\left(\frac{33}{38}\right)^2\right)y^2\leq221$$ or $$1\leq y\leq6.$$ The similar thing we can make for $x$. Now, check it.

Can you end it now?

I got $\{(2,5),(5,2)\}$.

Answer 2

A simpler approach:

$19y^2-(33x)y+19x^2-221=0$

we solve this for y:

$\Delta=(33x)^2-4\times 19\times (19x^2-221)$

Or:

$\Delta=16796-355\geq 0\Rightarrow x^2<47.31$

that is numbers $36, 25, 16, 9, 4, 1$ can be checked. Only For $x^2=25$ and $x^2=4$ we have integers for $\sqrt \Delta$ as 89 and 124 respectively which gives $x=5, y=2$. Since the equation is symmetric for x and y we also have $x=2, y=5$