<s>[INST] How to solve the problem on number theory Find the number of positive integer pairs $x,y$ such that 
$$xy+\dfrac{(x^3+y^3)}3=2007.$$
I solved the question by using factorization and further checking possible values of $x$ and $y$. But it was very lengthy as I had to check many cases for $x$ and $y$. Is there any possible other method?
 [/INST] Denote: $$\begin{cases}x+y=a \\ x-y=b\end{cases} \Rightarrow \begin{cases} x=\frac{a+b}{2}\\ y=\frac{a-b}{2}\end{cases}$$ Then:
$$xy+\dfrac{x^3+y^3}3=2007 \Rightarrow (a+2)^2+3b^2=\frac{24080}{a-1}.$$
Note that $a>1$ and $a$ can be $2,3, 5,6,8,9,11, 15,17,21$. Note that $a\ge 28 \Rightarrow b^2<0$. So, among them only $21$ suits. Hence $b=15$. Hence: $(x,y)=(18,3)$. Since the equation is symmetric, $(x,y)=(3,18)$. 
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