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Yesterday, at 23:18, I thought it was a remarkable moment of the day. The digits on the watch were providing a quadruplet of positive integers that satisfy the following system of equations: $$\begin{align} a^b &= cd \\ b^a &= c+d \end{align}$$

I wondered what the set of all positive integer solutions of this system was. Writing $d=b^a-c,$ I obtained a quadratic of $c$ with coefficients in terms of $a$ and $b$, which led me to the following: $$\left\{ c,d \right\} = \left\{\frac{b^a-\sqrt{b^{2a}-4a^b}}{2},\frac{b^a+\sqrt{b^{2a}-4a^b}}{2}\right\}$$

Therefore, given positive integers $a$ and $b,$ there is a solution if and only if $b^{2a}-4a^b$ is a perfect square. A brute force search using this result yields the following solutions: $(1,2,1,1),$ $(2,2,2,2),$ $(2,3,1,8),$ and $(2,3,8,1).$ I believe that there is no other solution than these. However, I'm unable to prove it. I would be glad if anyone could help me.

Fikilis
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    Exponential Diophantine equations are notoriously difficult. Do you have any reason other than numerical data to “believe” these are the only solutions? – Thomas Andrews May 19 '22 at 22:05
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    A generalization of Catalan's conjecture asserts that for every natural number $n$, there are only finitely many pairs of perfect powers with difference $n.$ And when $b$ is even, $b^{2a} - 4a^b$ is a difference of two perfect squares. Although $b$ doesn't have to be even, this conjecture combined with numerical data is what my instinct relies on. – Fikilis May 19 '22 at 22:22
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    For odd primes, a $p$-adic argument shows that if $p\mid c,d$ then $p^p\mid c,d$. – Servaes May 20 '22 at 02:11
  • Here is an idea. Ignore the case where all are even. All cannot be odd. Let $a$ and $c$ be even. We can move past the case where $a$ is a power of $2$. We get $a^b = 2^{kb} h^b$ where $h$ is odd and $k$ is positive integer. We also have $b^a = 2^{kb} + h^b$. With some effort, we can show that the RHS of last expression is less than LHS. – sku May 21 '22 at 08:23
  • For a further solution, one of $a,b$ must exceed $1000$. It would be very surprising if there would be another solution, but proving it is probably out of reach considering that Pillai's conjecture is only known for difference $1$. – Peter May 24 '22 at 08:58
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    Do you have an intetion to publish your conjecture? It's interesting, and my instinct shows that it may not be out of reach for some professionals out there. It is because proving Catalan's conjecture seems to be more difficult than proving yours. Or you may see some response at MO. – user1851281 May 26 '22 at 04:47
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    Using AM-GM inequality, we can see $\frac{b^a}{2} \geq \sqrt {a^b}$. I think our current tool of analysis can deal with this problem. – user1851281 May 26 '22 at 04:59
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    See this video. We get a lemma as IF $e<a<b$, THEN $a^b > b^a$. I think searching for some extent of number suffices to prove. – user1851281 May 26 '22 at 05:53
  • @user281 Using your idea, I've proved that if there is any other solution, it satisfies $3 \leq a < b$. However, it seems like this isn't enough to obtain a bound. Also, I'm not sure that this conjecture is significant enough to be published in a journal. What do you think? – Fikilis May 26 '22 at 16:55
  • @ErenKesim: Oh, no bound there? That's unfortunate. I think you can try it on "Monthly" since the idea seeems quite novel to me. I've seen many papers on Monthly which are less than 5 pages! Maybe including some codes and graphs from Python or R would be helpful for getting accepted! – user1851281 May 26 '22 at 19:55
  • @ErenKesim: Journal "Integer" also has some short papers with novel ideas. – user1851281 May 26 '22 at 20:21
  • @Fikilis I highly suspect that one can prove some upper bound on $b$ in the form that there are only finitely many solutions if $b \ge (2+\epsilon)a$ for any $\epsilon > 0$. – Ahmet Bilal Mar 10 '24 at 15:53
  • @Fikilis So like if we are able to prove that no solution exists for $a<b\le (2+\epsilon)a$, we will be left with checking cases where $a \le f(\epsilon)$ where $f$ is hopefully a very rapidly decreasing function. I numerically checked that $f(1)<3$. https://www.desmos.com/calculator/mnhvltimfn – Ahmet Bilal Mar 10 '24 at 16:42

1 Answers1

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Yesterday night I watched the video on youtube and tried my chance :) I was able to rule out some small number of cases.

If $b^{2a}-4a^b$ is a perfect square, then any $a>1$ must be even.

Proof. We use the already proven fact that $b^{2a}-4a^b$ is a perfect square iff there are positive integers $c,d$ satisfying the equations \begin{align} a^b &= cd \label{eq:cd} \tag{1} \\ b^a &= c+d \label{eq:c+d} \tag{2}. \end{align}

Assume that $b^{2a}-4a^b$ is a perfect square and $a > 1$ is an odd number. Then, $c$ and $d$ are odd numbers by \eqref{eq:cd}. But then $b$ must be even by \eqref{eq:c+d}. Let $b=2k$. Since $b^{2a}-4a^b$ is a perfect square, we have a Pythagorean triple of the form $p = (\star, 2a^k, (2k)^a)$. It is well-known that for any Pythagorean triple $(x,y,z)$ either all $x, y, z$ are even or $z$ is odd and $x$ and $y$ have different parity. Observe that simplifying the triple $p$ via dividing each term by 2 yields a new triple of the form $p' = (\star, a^k, 2^{a-1}k^a)$. No such Pythagorean triple can exist since $a^k$ is odd but $2^{a-1}k^a$ is not, a contradiction. $\tag*{\(\Box\)}$

OK, here is a follow-up result.

No solution exists for $b \ge 4a$.

Proof. First, we need the following simple lemma:

If $b^{2a} - 4a^b \ge 0$ and $b > 2a$, then $a<k^\frac{2}{k-2}$ where $k = b/a$.

Proof. By assumption there exists a real $k > 2$ such that $b=ka$. Then we have \begin{align} (ka)^{2a} &\ge 4a^{ka} \\ k^{2a}a^{2a} &> a^{ka} \\ k^{2a} &> a^{(k-2)a} \\ k^{2/(k-2)} &> a \end{align} $\tag*{\(\Box\)}$

It is obvious that all trivial solutions satisfy $b<4a$. It was proven that any other solution must satisfy $3 \le a < b$. Since we also proved that any $a>1$ must be even, we have $4 \le a < b$. Setting $b=4a$ in the lemma yields the bound $a<4$ for which there is no non-trivial solution. Note that $k^\frac{2}{k-2}$ is a decreasing function of $k$ for $k>2$. $\tag*{\(\Box\)}$

A simple brute-force search proves that any non-trivial solution must satisfy $1000 < a < b < 2.233a$.

There is no non-trivial solution if $b$ is an integer multiple of $a$. Specifically, the only possible case where $b=2a$ leads to $a=1, b=2$.

Proof. Let $b^{2a} - 4a^b = t^2$ for some positive integer $t$. Since $b=2a$, we have $(2a)^{2a} - 4a^{2a} = (2^{2a}-4)a^{2a} = t^2$. Observe that $a^{2a}$ is a square number. Thus, $2^{2a}-4$ must be a square as well. But $2^{2a}$ is also a square. There is a single pair of squares whose difference is $4$, which is $(0,4)$. Therefore, we have $a=1$. $\tag*{\(\Box\)}$

Ahmet Bilal
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  • İspata ulaşmaya gerçekten yaklaştık gibi duruyor, detayları özelden konuşmak isterim. Matematikçi misin, nerede okuyor ya da çalışıyorsun? Sana nasıl ulaşabilirim? – Fikilis Mar 15 '24 at 14:41
  • @Fikilis telegram id neutralsecond discord kullanıcı adı da aynı – Ahmet Bilal Mar 15 '24 at 16:13
  • With similar methods, we can also prove that $\frac{\ln(b)}{\ln(a)}$ is close to 1. – Fikilis Mar 15 '24 at 16:48