All pairs (x,y) that satisfy the equation $xy+(x^3+y^3)/3=2007$

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

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

• Are you familiar with the factorization of $a^3 + b^3 + c^3 - 3abc$? (This is not a terribly well-known factorization.) Commented Aug 30, 2012 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? Commented Aug 30, 2012 at 15:54
• @Qiaochu Yuan can you elaborate a bit more , please ?
– user373141
Commented Mar 26, 2017 at 17:03

Observe that the equation is symmetric.

As $$3\mid(x^3+y^3)$$, 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 $$(1,6)$$.

• 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 $12$. Commented Aug 30, 2012 at 12:30
• @ErickWong, thanks for your instructive feedback. Commented Aug 30, 2012 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.

• can you elaborate a bit more please ?
– user373141
Commented Mar 26, 2017 at 17:05
• use modular arithmatic Commented May 31, 2019 at 8:10

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 $$(3,18)$$