How to solve the problem on number theory

Question

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?

Answer 1

Another possible way:

• multiply the equation by $3 $.
• show that $ x =-y\ \bmod 3 $ using FLT.
• rule out the cases $1 $ and $2 $, so that $3\mid x$
• the equation is now $ kl+k^3+l^3=223 $.
• perfect cubes close enough to $223 $ are few, one easily checks that $(6, 1) $ and $(1, 6) $ are the only solutions as you found. Congrats.

Answer 2

Idea:

$$ 3xy+(x+y)(x^2-xy+y^2)=6021$$ so $$ 3xy+(x+y)(x+y)^2-2xy(x+y)=6021$$ Now let $a=x+y$ and $b=xy$. Then $$ a^3-6021=3b(a-1)$$ so $3\mid a$ so $a=3c$ and know we have $$9(c^3-223) =b(3c-1)$$ so $9\mid b$ so $b=9d$ and thus $$c^3-223 =d(3c-1)\implies 3c-1\mid c^3-223$$

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I thought that this with $c$ will go faster than with $a$, so I deleted it. Anway here is what I have before:

so $$a-1\mid a^3-6021\;\;\;\;{\rm in}\;\;\;\; 3\mid a$$ Since $a-1\mid a^3-1$ we have $$a-1\mid 6020\implies $$ $$a-1\in \{1,2,4,5,7,10,14,20,28,35, 43,70,86,140,172,215,301,430, 602,860,1202,1505, 3010,6020\}$$ so $$a\in \{2,3,5,6,8,11,15,21,29,36, 44,71,87,141,173,216,302,431, 603,861,1203,1506, 3011,6021\}$$ so

$$a\in \{3,6,15,21,36,87,141,216,603,861,1203,1506,6021\}$$

Answer 3

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, $(x,y)=(3,18)$.

Answer 4

The first thing that screams at me is $3|x^3 + x^3$ so

$x^3 \equiv -y^3 \mod 3$ so $x\equiv -y \mod 3$

Let $x \equiv i \mod 3$ and $y \equiv -i \mod 3$. If $x = 3k + i$ and $y = 3j -i$ then $\frac {x^3 + y^3}{3} = 9(k^3 + j^k) + 9(k^2i-j^2i)+3(ki -ji) + \frac {i^3 - i^3}3 \equiv 0 \mod 3$

So $xy + \frac {x^3 + y^3}3 \equiv i^2 \mod 3$ and $2007\equiv 0 \mod 3$ so $i = 0$ and $3|x$ and $3|y$.

Let $x = 3a$ and $y = 3b$ and we get

$9ab + 9(a^3 + b^3) = 2007$ so

$ab + a^3 + b^3 = 223$

Hmmm, still trial and error but $b^3, a^3 < \sqrt[3]223 \approx 6$ so not too many to test.

One of $a$ or $b$ must be odd. We have $ab + a^3 + b^3\equiv ab + a +b \equiv 1 \mod 3$: That means:

$(a,b)\equiv (0, k) \to 0*k + 0 + k\equiv k \mod 3$

So $a\equiv 0$ (wolog) and $b \equiv 1$ is possible.

$(a,b) \equiv (1,1) \to 1 + 1 + 1 \equiv 0 \mod 3$.

$(a,b) \equiv (1,-1) \to -1 + 1 + -1 \equiv -1 \mod 3$.

$(a,b) \equiv (-1,-1) \to 1 -1 -1 \equiv -1 \mod 3$.

So $a \equiv 0$ and $b \equiv 1$ and at least one is odd.

So

$[0,1],[3,1],[3,4],[6,1]$ are the only four options.

Obviously if the average of $ab, a^3, b^3$ is $\frac {223}3 \approx 70 > 4^3$, the first three can't possibly work. Don't even need to test them.

So $[6,1]$ is only option.

An indeed $6 + 6^3 + 1 = 223$.

So, if my reasoning is right (which.... I really don't see any error or faulty assumptions... but ... a self-editor is always blind....). $\{x,y\} = \{18,3\}$ are the only solutions.

Answer 5

I show here that the only correct answer is, as it maintain @fleablood and @farruhota in their answers. $$\frac{x^3+y^3}{3}=\frac{(x+y)^3}{3}-xy(x+y)\Rightarrow xy+\frac{(x+y)^3}{3}-xy(x+y)=2007$$ It follows $x+y\equiv 0\pmod3$. The cubic $$xy+\frac{x^3+y^3}{3}=2007$$ is symmetric respect to the diagonal $y=x$ then if the line $x + y = 3a$ cuts the cubic at the point $(x, y)$ it also cuts it at the point $(y, x)$. The system $$\begin{cases}x+y=3a\\xy+\dfrac{x^3+y^3}{3}=2007\end{cases}$$ has a solving equation of the second degree whose solution $(x,y)$ (and the symmetric $(y,x))$ is$$\begin{cases}x=\dfrac{3(-3a^2+a+X)}{2-6a}\\y=\dfrac{3(3a^2-+Xa+X)}{2-6a}\end{cases}\quad\text{ where } X=\sqrt{-3a^4-2a^3+a^2+2676a-892}$$ It follows that there are quite few values of $a$ giving real roots. In the attached graph of the cubic, only the black lines $x + y = 21,24,27$ (corresponding to $a=7,8,9$) cut the curve in real positive points. For $a\le6$ and $a\ge10$ does not. For the line $x+y=18$ the red points of intersection are not in the first quadrant (approximately $(18.204,-0.204)$).

Finally we have just three candidates (and the symmetric) produced by the lines $x+y=21,24,27$ but only the line $x+y=21$ agrees with the requirement of positive integers for the intersection. In this line we have marked all the points of integer coordinates of the first quadrant (in red those of intersection with the cubic). These three candidates are $$(18,3)\quad\text{ and approximately }\quad(17.56,6.44),(16.162,10.838).$$ Thus the only solutions are given by $\{18,3\}$.