How to solve the Diophantine equation: $x^3 + y^3 + xy = 223$ for all integer pairs $(x,y)$?

Question

I have tried factorizing, inequality bounds and any standard technique I could, but somehow nothing seems to work. I know one solution is $(1,6)$ and $(6,1)$, but that was through trial and error. How to do it "mathematically"?

Answer 1

If $x=0$ or $y=0$ it’s straightforward (no solutions).

If $x,y> 0$ it’s easy to simply search for solutions ($x^3 \leq 223$ so $x \leq 6$, same for $y$).

If $x,y < 0$, it’s easy enough as well: write $x=-a$, $y=-b$ with $a,b>0$, then $a^3+b^3-ab=-223$ thus $ab > a^3+b^3 \geq a^2+b^2$, contradiction.

If $y=-b, x=a$ with $a,b>0$ (by symmetry, the only remaining case), then the equation becomes $a^3=223+b^3+ab$, so $a > b$. Write $a=c+b$, then $3cb^2+3c^2b+c^3=223+b^2+cb$.

Because $2cb^2 \geq b^2+cb$, it follows that $cb^2+3c^2b+c^3 \leq 223$, in particular $c \leq 6$ and $b \leq 14$. Then you can brute force again for solutions.

Answer 2

Let $s=x+y$ then $$x^3 + (s-x)^3 +x(s-x)-223=0\Leftrightarrow (3s-1)x^2-s(3s-1)x+(s^3-223)=0.$$ The quadratic equation with respect to $x$ has discriminant: $$\Delta=-(3 s-1)(s^3 + s^2 - 892)$$ which is $\geq 0$ just when the integer $s$ belongs to the interval $[1,9]$ (notice that the sign of the second factor is easy to find since its derivative is $s(3s+2)$).

Looking in that interval for perfect squares, we find just one value for $s$, i.e. $7$ with $\Delta=100^2$. Then $$x=\frac{s(3s-1)\pm\sqrt{\Delta}}{2(3s-1)}=\frac{7\cdot 20 \pm100}{2\cdot 20}=1,6.$$ Hence the integer solutions $(x,y)$ are just $(1,6)$ and $(6,1)$.

Answer 3

Number theoretic solution.

Let $x+y=u$, $xy=v$. Then $u^3-3uv+v=223$, which solves out $v=\dfrac{u^3-223}{3u-1}$. $9v\in\mathbb Z$ implies $3u-1\mid27u^3-6021$. Since $$3u-1\mid(3u)^3-1^3=27u^3-1,$$ also $3u-1\mid6020$.

Let $3u-1=k$ to have $u=\dfrac{k+1}3$, $v=\dfrac{\left(\frac{k+1}3\right)^3-223}k$. Notice that $u^2-4v=(x-y)^2\ge0$. By doing simple approximation, this doesn't hold for $k<0$ and starts to fail when $k\ge29$. So we only look at the small factors and get $$\begin{array}{|l|l|}\hline u&v\\\hline\hline 1 & -111 \\ 2 & -43 \\ 5 & -7 \\ 7 & 6 \\\hline \end{array}$$ By checking whether $u^2-4v$ is a perfect sqaure, we find $(7,6)$ is the only possibility. Since $\begin{cases}x+y=7;\\xy=6.\end{cases}$ imply $(x,y)=(1,6)$ or $(6,1)$, they're our only solutions.

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Full results of $(u,v)$ by python listed below. But using my solution above you don't need software at all. $$\begin{array}{|l|l|}\hline u&v\\\hline\hline 0 & 223 \\ 1 & -111 \\ -1 & 56 \\ 2 & -43 \\ -2 & 33 \\ -3 & 25 \\ 5 & -7 \\ 7 & 6 \\ -9 & 34 \\ 12 & 43 \\ -14 & 69 \\ -23 & 177 \\ 29 & 281 \\ 47 & 740 \\ -57 & 1078 \\ 72 & 1735 \\ -100 & 3323 \\ -143 & 6801 \\ 201 & 13489 \\ 287 & 27488 \\ -401 & 53556 \\ 502 & 84057 \\ -1003 & 335225 \\ 2007 & 1342906 \\\hline \end{array}$$

Answer 4

I was working on another solution (finally got it!), so thought it would be worthwhile to post for others to comment and give feedback.

In the equation, $a^3 + b^3 + ab = 223$ multiply both sides by 27. This gives $(3a)^3 + (3b)^3 + 3\cdot(3a)\cdot(3b) = 6021.$

Now, let $3a = x, 3b = y$.

The equation now becomes: $x^3 + y^3 + (-1)^3 + 3\cdot(x)\cdot(y)\cdot(-1) = 6020$

Using the identity $x^3 + y^3 + z^3 - 3xyz = (x+y+z)(x^2 + y^2 + z^2 - xy- yz - zx)$,we can write the above equation this way:-

$(x + y − 1)((x + y)^2 + x + y + 1 − 3xy) = 6020 = 2^2\cdot5\cdot7\cdot43$

We can pretty much brute force our way to the answer now, and substitute $x$ for $\frac{a}{3}$ to get $(6,1)$ and $(1,6)$.