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!
Best Answer
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:
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.