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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.)2012-08-30
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    @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?2012-08-30
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    @Qiaochu Yuan can you elaborate a bit more , please ?2017-03-26

3 Answers 3

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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 $12$.2012-08-30
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    @ErickWong, thanks for your instructive feedback.2012-08-30
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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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    can you elaborate a bit more please ?2017-03-26
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Note that the given equation is $$x^3+y^3+3xy=6021$$ or $$x^3+y^3-1+3xy=6020$$ Factoring it we get , $$(x+y-1)(x^2+y^2+1+x+y-xy)=2^2.5.7.43$$ Obviously check $\equiv 3$ and see $$x+y-1\equiv 2 \mod 3$$ Also, $$x+y-1 < x^2+y^2+1+x+y-xy$$ This means, $$x+y-1 \rightarrow 20,5,2,35$$ so now it's easy to see that $$x+y-1=20$$ and so $(x, y)=(18,3)$ or $(3,18)$