Solve over natural numbers: $m^3=2n^3+6n^2$. Functional equation gives rise to a diophantine equation!

Question

My question is basically to find all natural numbers $(m,n)$ such that

$m^3=2n^3+6n^2$

First for some background (this is not really that relevant but anyways): I was trying to solve an olympiad functional equation

$f:\mathbb{N} \to \mathbb{N}$ such that $f(x+y)=f(x)+f(y)+3(x+y)\sqrt[3]{f(x)f(y)}$

Then we will get that $f(x)$ will always be a perfect cube so setting $f(x)=g(x)^3$ and then $P(1,1)$ gives $g(2)^3=2g(1)^3+6g(1)^2$. And then I noticed that if $g(1)=1$ we could just induct up to get $g(x)=x$ always, so this is where the question comes from, because if i prove the only solution in natural numbers to the diophantine in $(1,2)$ I will be done.

Now back to the diophantine, obvioulsy $(m,n)=(2,1)$ is a solution and wolfram alhpha says it is the only solution. For the past one hour or so I have been trying to prove this but have done basically nothing...

I have just been doing random subsitutions and making cases and stuff, like $m=2y$ so then we get $y \equiv x \mod 3$

and then making cases like if $x$ is divisible by $3$ or not but I have not even been able to rule out one case...

There are infact more integer solutions so I believe we could use some inequalities somehow but I am not sure how whatsoever...

Any help on this problem will be appreciated...

Thanks!

Answer 1

The following will be a general way to handle questions of the sort $|x^3 -a y^3| \le b$ for small values of $a,b$:

In this answer, it was reduced to:

In summary, we need to solve the equations \begin{eqnarray} X^3-2Y^3&=&1,\\ X^3-4Y^3&=&1,\\ X^3-2Y^3&=&3,\\ X^3-4Y^3&=&3, \end{eqnarray} over the integers.

I will find all solutions to $|x^3 - 2 y^3| \le 6$. This is enough since each of the equations can be changed into this form (for example: $x^3 - 4y^3 = 3 \iff (2y)^3 - 2 x^3 = -6$.

I will use methods of paper [A] (cf. Lemma 3 and Lemma 4 there) and instead of the bound of Easton used in [A] I will use bounds given in [B]. I will also remark that there's a quicker proof directly using Theorem 6.1 of [B], but it leaves us a few more numbers to check (one gets $|x| \le 36$), and I wanted to keep the bounds small enough to easily check by hand.

We will prove Lemma 3 and Lemma 4 of [A] with better bounds as follows:

Lemma 3: If a pair of integers $(x,y)$ satisfies $|x^3 - 2y^3|\le 6$, then either $|y|\le 2$ or $\frac xy$ is a convergent in the continued fraction expansion of $\sqrt[3]{2}$.

Proof: Suppose $|x^3 - 2y^3|\le 6$ and $|y|\ge 3$. It follows that $$\left|\frac{6}{y^3} \right| \ge \left|\frac{x^3}{y^3} - 2 \right| = \left| \frac xy - \sqrt[3]2 \right| \left(\frac{x^2}{y^2} + \sqrt[3]2 \frac xy + \sqrt[3]4 \right).$$ Since $\left|\frac{6}{y^3} \right| \le \frac{6}{27} = 0.222$, we have $2 - (x/y)^3 \le 0.222$ and so $x/y > 1.21$; hence the inequality $$\left(\frac{x^2}{y^2} + \sqrt[3]2 \frac xy + \sqrt[3]4 \right) > 4.5$$ This implies that $$\left| \frac xy - \sqrt[3]2 \right| < \frac{6}{4.5} \cdot \frac{1}{|y|^3} = \frac{4}{3} \cdot \frac{1}{|y|^3} -(*)$$ Now, for $|y| \ge 3$, $$\frac{4}{3} \cdot \frac{1}{|y|^3} < \frac 12 \cdot \frac{1}{|y|^2}$$ and it is well known that any $x/y$ satisfying $\left| \frac xy - \sqrt[3]2 \right| < 1/(2y^2)$ is a convergent. $\square$

Lemma 4: Suppose that a pair of integers $(x,y)$ satisfy $|x^3 - 2y^3| \le 6$ and $|y| \ge 3$. Then one has $|y|\le 28 $.

Proof: [B] gives us that for integers $x,y$, the inequality $$\left| \frac xy - \sqrt[3]2 \right| > \frac 14 \cdot \frac 1{y^{2.5}}$$ holds. Combining this with the inequality $(*)$ derived in the previous lemma we obtain $$y^{1/2} < \frac {16}3 \implies y \le 28. \square$$

Now, the continued fraction expansion of $\sqrt[3]{2}$ begins as $[1,3,1,5,1,1, \ldots]$. So, the convergents are $p_n/q_n$, with $$q_0 =1, q_1 = 3, q_2 = 4, q_3 = 23, q_4 = 27, q_5 > 28.$$ Also, $p_0 =1, p_1 = 4, p_2 = 5, p_3 = 29, p_4 = 34$.

So, we need to check these convergents and $y = 0, \pm 1, \pm 2$. We obtain: $(5,4), (0,0), (\pm 1,0), (\pm 1, \pm 1), (2, 1), (-2,-1)$.

I will leave the verification that this means that $(m,n) = (2,1)$ is the only solution of $m^3 = 2n^3 + 6n^2$ up to you.

As a final remark, the way that these methods generalize to $|x^3 - ay^3| \le b$ is because Lemma 3 works quite generally for any value of $a,b$, and Lemma 4 works for small values of $a$ due to the bounds given in Corollary 1.2 of [B].

[A] - F. Beukers and J. Top, 'On oranges and integral points on certain plane cubic curves', Nieuw Arch. Wisk. (4) 6 (1988), 203-210

[B] - Bennett, Michael A.. “Effective measures of irrationality for certain algebraic numbers.” Journal of the Australian Mathematical Society. Series A. Pure Mathematics and Statistics 62 (1997): 329 - 344.

Answer 2

This doesn't answer the diophantine equation $ m^3 = 2n^3 + 6n^2$ (which is hard), but answers the functional equation $f(x+y) = f(x) + f(y) + 3(x+y) \sqrt[3]{f(x) f(y) } $ when defined over all integers $ f: \mathbb{Z} \rightarrow \mathbb{Z}$.

With $ x = y = 0$, $f(0) = 0$.
Now, suppose there is some $ a \neq 0 $ such that $f(a) = 0$, then with $ y = a$, we get that $f(x+a) = f(x)$ , so we have a periodic function. If there is some $x$ with $f(x) \neq 0$, then $f(x+x+a) = f(x+x)$ gives us $ 3(x+x+a)\sqrt[3]{f(x)f(x) } = 3(x+x)\sqrt[3]{f(x) f(y)}$, which is a contradiction. Hence, the only solution is $f(x) = 0 \, \forall x$ (which doesn't give us a solution on the positive integers). Otherwise, $f(x) \neq 0 \Leftrightarrow x \neq 0$.

We will show that $\forall x, y \neq 0$, $$\frac{f(x) } { x^3 } = \frac{f(y) } { y^3}.$$ This is obvious if $ x = - y$ by definition.
Otherwise, let $ z = - x - y \neq 0$. So $-f(x) = f(y+z) = f(y) + f(z) - 3x \sqrt[3]{f(y) f(z) }$, or that $ f(x) + f(y) + f(z) = 3x \sqrt[3]{ f(y) f(z) } $.
Swapping $x$ and $y$, we get that $ f(y) + f(x) + f(z) = 3y \sqrt[3]{f(x) f(z) }$.
Hence, $ \frac{f(x) } { x^3 } = \frac{f(y) } { y^3}$ as claimed. This has value $f(1)$.

Finally, setting $f(x) = f(1)x^3$ with $f(1) \neq 0$ in the functional equation, we get $$f(1) ( x+y)^3 = f(1) x^3 + f(1) y^3 + 3(x+y) f(1)^{2/3} xy$$ or that $3xy(x+y) ( f(1) - f(1) ^ { 2/3} ) = 0 $.
This gives us $ f(1) = 1$ so $ f(x) = x^3 \, \forall x \in \mathbb{N}$.

Note

• Instead of solving $ m^3 = 2n^3 + 6n^2$, we're solving $ 8n^3 = 2n^3 + 6 n^2$ which is much easier.
• We might be able to work around extending the function to the negative integers (EG replacing $f(z) = f(-x-y) = - f(x+y)$), but that solution presentation felt the most natural to me.
• In fact, this solution also works if the domain and range are the reals. We didn't use any facts about integers / number theory.

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Ignore this:

Suppose that the function is only defined for the positive integers.
We extend the function to the non-negative integers by defining $f(0) = 0$.
It remains to check that for $ x, y \geq 0$, we still have $ f(x+y) = f(x) + f(y) + 3(x+y) \sqrt[3]{f(x)f(y)}$, which is obvious in the cases of $ x = 0 , y = 0 , x = y = 0$.
Again, we extend the function to the negative integers by defining $f(-x) = - f(x)$. We have to check that the functional equation holds, which is left as an exercise to the reader.
Of course, we now have a superset of solutions, and have to check which allow for solutions to the positive integers.

Answer 3

The attempted answer below shows that it suffices to solve four diophantine equations of the form $$X^3-aY^3=b,$$ with $a\in\{2,4\}$ and $b\in\{1,3\}$. After user Sil pointed out some gaps in my reduction, I have completed it, but it has become quite a cumbersome mess. My attempt to reduce the mess leads to my answer becoming identical to Sil's, so I leave it here as is.

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Let $m$ and $n$ be natural numbers such that $$m^3=2n^3+6n^2.$$ First note that $(m,n)=(0,0)$ is a solution, and that every other solution has $m,n>0$. The right hand side factors as $$m^3=2n^2(n+3),$$ which shows that $m$ is even and that $n^2$ divides $m^3$. Let $m=2k$ so that $$4k^3=n^2(n+3).$$ Let $p>3$ be a prime number dividing $n$, and let $a$ be the greatest power of $p$ dividing $n$. Then $p^{2a}$ divides $n^2$, and $p$ does not divides $n+3$. It follows that the greatest power of $p$ dividing $4k^3=n^2(n+3)$ is $p^{2a}$. Because $p$ does not divide $4$, this shows that the greatest power of $p$ dividing $k^3$ is $p^{2a}$, meaning that $3$ divides $a$.

This is true for every prime $p>3$, and so we see that $n=2^b3^cN^3$ for some positive integer $N$ coprime to $2$ and $3$, and similarly that $k=2^d3^eK^3$ for some positive integer $K$ coprime to $2$ and $3$.

If $c>0$ it follows that $$2^{3d+2}3^{3e}K^3=4k^3=n^2(n+3)=2^{2b}3^{2c+1}N^6(2^b3^{c-1}N^3+1).$$ We see that $N^6$ divides $K^3$ and so $K=LN^2$ for some positive integer $L$. Then $$2^{3d+2}3^{3e}L^3=2^{2b}3^{2c+1}(2^b3^{c-1}N^3+1).$$ If $b>0$ and $c>1$ then the factor $2^b3^{c-1}N^3+1$ is coprime to $2$ and $3$, and so we find that $3d+2=2b$ and $3e=2c+1$. In particular $b\equiv c\equiv1\pmod{3}$, say $b=3B+1$ and $c=3C+1$. Then $$L^3=2^{3B+1}3^{3C}N^3+1=2(2^B3^CN)^3+1.$$ This 'reduces' the problem to finding positive integer solutions to $$X^3-2Y^3=1.$$ It is a nontrivial result that this has no notrivial solutions (I believe it is in Mordell, but it escapes me at the moment).

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As noted in the comment by Sil, this leaves the edge cases where either $b=0$ or $c\leq1$.

If $c=0$ then also $e=0$, and so we get $$2^{3d+2}K^3=2^{2b}N^6(2^bN^3+3).$$ Again $K=LN^2$ for some positive integer $L$ and so $$2^{3d+2}L^3=2^{2b}(2^bN^3+3).$$ If $b>0$ the we see that $3d+2=2b$ and so we are left with solving $$X^3-2^fY^3=3,$$ for $f\in\{0,1,2\}$. So this boils down to solving $$X^3-2Y^3=3\qquad\text{ and }\qquad X^3-4Y^3=3.$$ If $b=0$ we get $$4(2^dL)^3=N^3+3,$$ and so we are again left with solving $X^3-4Y^3=3$ over the integers.

If $c=1$ and $b>0$ then the original argument still shows that $$2^{3d+2}3^{3e}L^3=2^{2b}3^3(2^bN^3+1),$$ where the factor $2^bN^3+1$ is coprime to $2$, and so $3d+2=2b$. Then still $b=3B+1$ for some integer $B$, and so we find that $$(3^{e-1}L)^3=2(2^BN)^3+1,$$ and we are again left with solving $X^3-2Y^3=1$ over the integers.

Finally, if $c=1$ and $b=0$ then the argument above still shows that $$2^{3d+2}3^{3e}L^3=3^3(N^3+1).$$ It follows that $$4(2^d3^{e-1}L)^3=N^3+1,$$ and so we are again left with solving $X^3-4Y^3=1$ over the integers.

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In summary, we need to solve the equations \begin{eqnarray} X^3-2Y^3&=&1,\\ X^3-4Y^3&=&1,\\ X^3-2Y^3&=&3,\\ X^3-4Y^3&=&3, \end{eqnarray} over the integers.

Answer 4

This is not a complete solution but it is too long for a comment.

Let $g=(m,n)$ and $m=ag,n=bg$ with $(a,b)=1$. Then $a^3g=2b^2(bg+3)$ and so $b^2\mid g$, giving $g=kb^2$, transforming the equation eventually to $$k(a^3-2b^3)=6.$$ This gives four equations $a^3-2b^3=c$ where $c\in \{1,2,3,6\}$.

Cases $a^3-2b^3=1$ and $a^3-2b^3=2$ can be both resolved using Skolem's theorem (see also this answer ):

For any integer $d \ne 0$, there exists at most one pair $(x,y) \in \mathbb{Z} \times \mathbb{Z}$ with $y \ne 0$ such that $$x^3 + dy^3 = 1$$

We can see that for $d=2$ such non-trivial solution is $(x,y)=(-1,1)$ and for $d=4$ the linked resource shows there is no non-trivial solution.

For $a^3-2b^3=1$ using the above (and writing $a^3+2(-b)^3=1$) we get only solutions $(-1,-1)$ and $(1,0)$. For $a^3-2b^3=2$ we let $a=2u$ and the equation translates to $4u^3-b^3=1$, which fits the above theorem as $(-b)^3+4u^3=1$. So using the above result again, we have $(b,u)=(-1,0)$ as the only solution. None of these solutions correspond to a positive solution of the original problem.

So it "only" remains to solve $a^3-2b^3=3$ and $a^3-2b^3=6$ (which can be transformed to $a^3-4b^3=3$ using that $a$ must be even). Note that former has solutions $(-5,-4)$ and $(1,-1)$ which are not positive, while the latter has a solution $(a,b)=(2,1)$, which corresponds to the solution noted in the OP. Whether these are the only solutions, I do not know.