# positive integer solutions to $x^3+y^3=3^z$

I am seeking all positive integer solutions to the equation $$x^3+y^3=3^z$$.

After doing number crunching, I think there are no solutions. But I am unable to prove it.

Attempt

If $$x$$ and $$y$$ have common divisor $$d$$, we have $$d^3(m^3+n^3)=3^z$$. So $$d$$ must be a power of $$3$$, and we are back to where we started. So we assume $$x$$ and $$y$$ are coprime.

Testing the parity, we have sum of 2 cubes to be odd. WLOG, we can assume $$x$$ is even and $$y$$ is odd.

Trying mod $$3$$, we have $$x+y=0 \pmod 3$$. Since $$x$$ and $$y$$ are coprime, $$x$$ and $$y$$ must be congruent to $$1$$ and $$-1$$ or vice-versa.

If I assume $$x=3m+1$$ and $$y=3n-1$$, expand out and simplify, I get $$27(m^3+n^3)+27(m^2-n^2)+9(m+n)=3^z$$. If I assume $$z \geq 3$$, this gives $$(m^3+n^3)+(m^2-n^2)+\frac{m+n}{3}=3^{z-3}$$. But I don't see how to proceed.

I also tried mod $$9$$ but didn't get anywhere, it didn't cut down the possibilities by much.

I also tried letting $$y=x+r$$. Then \begin{align*} x^3+y^3 &= x^3+(x+r)^3 \\ &= x^3 + (x^3+3x^2r+3xr^2+r^3) \\ &= 2x^3+3x^2r+3xr^2+r^3 \\ &= 3^z \end{align*}

Then $$3\mid 2x^3+3x^2r+3xr^2+r^3$$, and $$3\mid 3x^2r+3xr^2$$, so this implies $$3 \mid 2x^3+r^3$$. But this doesn't yield any contradiction.

Can anyone supply a proof? Or if my hypothesis is wrong, how to derive all the integer solutions?

Thank you.

• Maybe concentrate on $z$? It can’t be a multiple of $3$, due to Fermat. Sep 9, 2018 at 9:20
• Are you familiar with the Lifting Exponent Lemma? Thus is a classic example of this. I think that there are an article in www.mathlinks.ro Sep 9, 2018 at 9:37
• @RicardoLargaespada Can we apply it here? the LTE requires $p \mid x-y$ Sep 9, 2018 at 10:28
• Also you have to read Zsigmondy Theorem, the solution by this technique is awesome Sep 9, 2018 at 15:52
• I'm confused; isn't $(x,y,z)=(1,2,2)$ a solution? Nov 19, 2020 at 13:35

$27(m^3+n^3) +27(m^2-n^2) + 9(m+n) = 3^z$

$(m+n)(3m^2 +3n^2-3mn+3m-3n+1) = 3^{z-2}$

since $z \ne 2$, we have $3(m^2 +n^2-mn+m-n)+1$ divides a power of $3$ or is equal to $1$, both of which is not possible.

• Now that you write it out, it seems really obvious XD. Thanks! Sep 9, 2018 at 12:10
• You are welcome :) Sep 10, 2018 at 17:49
• But it is equal to $1$ for $(m,n)=(0,1)$. And why "...since $z\neq2$..."? It would make more sense to infer that $$3(m^2 +n^2-mn+m-n)+1,$$ must be a power of $3$, which happens only if it equals $1$, and from here conclude that $z=2$. Nov 19, 2020 at 13:39

You already proved that we can assume that $x,y$ are relatively prime.

We can find easly solution for $n\leq 2$. Let $n\geq 3$.

So $x+y = 3^a$ and $x^2-xy+y^2=3^b$ for some nonegative integers $a+b=n$.

If $b\geq 2$ then $9\mid x^2-xy+y^2$ and $3\mid x+y$, so $$9\mid (x+y)^2-(x^2-xy+y^2)=3xy\implies 3\mid xy$$ A contradiction. So $b\leq 2$ and this should be easly done.

• that is really an ingenious way to show it. Thanks! Sep 9, 2018 at 12:10

Solution by LTE, let $d=(x,y)>1$, $x=dx_1$, $y=dy_1$, $(x_1,y_q)=1$ then $$d^3\mid x^3+y^3=3^z$$ then $d\mid 3^a$ for a positive integer $a<z/3$. Dividing by $d^3$ we obtain $x^3_1+y^3_1=3^b$, for a positive integer $b=z-a$. This is a equation similar to the original, then we can assume $(x,y)=1$. If $3\nmid x+y$ then $x+y=1$ a contradiction. Then we have all the conditions to use LTE: $$v_3(x^3+y^3)=v_3(x+y)+v_3(3)=v_3(x+y) +1=v_3(3^z)=z$$ From here we conclude that $x+y=3^{z-1}$ this implies that $x^2-xy+y^2=3$, $(x-y)^2+xy=3$, if $|x-y|\ge 2$ we have a contradiction, then there are two cases: Case 1. $|x-y|=0$ this is not possible due $x$ and $y$ are coprimes and $x+y>2$. Case 2. $|x-y|=1$ implying $xy=2$, with solutions $(x,y)=(2,1)$ or $(1,2)$, implying $3^{z-1}=2+1\Rightarrow z=2$.

Then the solutions to the original equation are: $(x,y,z)=(3^k,2\cdot 3^k,3k+2)$ or $(2\cdot 3^k, 3^k,3k+2)\forall k\in \mathbb{N}$.