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How we can find the all pairs $(x,y)$ from the integers numbers ,that satisfy the equation :

$$xy+\frac{x^3+y^3}{3} =2007$$

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Are you familiar with the factorization of $a^3 + b^3 + c^3 - 3abc$? (This is not a terribly well-known factorization.) – Qiaochu Yuan Aug 30 '12 at 8:09
@QiaochuYuan, are you saying $x^3+y^3+(-1)^3-3xy(-1)=(x+y-1)(x^2+y^2+1-xy+x+y)$? Could you please elaborate a bit? – lab bhattacharjee Aug 30 '12 at 15:54

Observe that the equation is symmetric.

As $3|(x^3+y^3)$, either $(x,y)$ will be $(3a+1,3b-1)$, $(3a-1,3b+1)$ or $(3a,3b)$.

If $(x,y)$ is $(3a+1,3b-1)$, $\frac{x^3+y^3}{3}=3(3a^3+3b^3+3a^2-3b^2+a+b)$

So, 3 must divide $xy$ which is impossible as $xy=(3a+1)(3b-1)$

So, $(x,y)$ will be $(3a,3b)$.

So,$9(ab+a^3+b^3)=2007\implies a^3+b^3+ab=223$

Now, 223 is prime, so, $(a,b)=1$

If we think of solution in natural number, $a<7$ .

By trial (which is aided by $(a,b)=1$), $(a,b)$ is $(6,1)$ or $(1,6)$.

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The question specifies integers, not just natural numbers. However, it is not hard to see that at least one of $a,b$ must be positive. If $a > 0 > b$ then $a^3 + b^3 + ab$ takes its least positive value when $|b|=a-1$. This still constrains $a$ to be less than $12$. – Erick Wong Aug 30 '12 at 12:30
@ErickWong, thanks for your instructive feedback. – lab bhattacharjee Aug 30 '12 at 15:52

Let $x+y=a,xy=b$ then the equation is equivalent to $$a^3-3abc+3b=6021$$ or $$(a-1)(a^2+a+1-3b)=6020$$ Now it is easy to do.

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