# Positive integers satisfying $a^b = cd$ and $b^a = c+d$

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.

• Exponential Diophantine equations are notoriously difficult. Do you have any reason other than numerical data to “believe” these are the only solutions? Commented May 19, 2022 at 22:05
• 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. Commented May 19, 2022 at 22:22
• For odd primes, a $p$-adic argument shows that if $p\mid c,d$ then $p^p\mid c,d$. Commented May 20, 2022 at 2:11
• 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. Commented May 26, 2022 at 4:47
• 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. Commented May 26, 2022 at 4:59

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 $$a where $$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$$}$$

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