Find all solutions in positive integers to $a^b -b^a = 3$

Question

Find all solutions in positive integers to $a^b -b^a = 3$

It appears that the only solution is $a=4, b=1$.

Modular arithmetic is not of much help (apart from deriving that $a,b$ must have opposite parity).

We know that $a^x$ dominates the polynomial $x^a$ but this does not seem to yield the result.

Tried hard to estimate a bound for $a^b - b^a$ but with no success.

Any help is appreciated. This is a problem from a regional mathematics Olympiad.

Added on 26 March, 2017:

I was trying to find a solution that involves no Calculus (or a minimum of Calculus) since this is a problem from a Junior Olympiad in which the students do not have a knowledge of Calculus. The following argument seems to work. Please point out flaws/mistakes, if any.

Lemma 1 For any $n \geq 4$, $$n^{n+1} - (n+1)^n > (n-1)^n - n^{n-1}$$ Proof \begin{align*} &\qquad n^{n+1} - (n+1)^n > (n-1)^n - n^{n-1} \\ &\Leftrightarrow n^{n+1} + n^{n-1} > (n-1)^n + (n+1)^n \\ &\Leftrightarrow n+\frac{1}{n} > \left(1-\frac{1}{n}\right)^n + \left(1+\frac{1}{n}\right)^n \end{align*} Since $\left(1+\frac{1}{n}\right)^n < 3$ and $\left(1-\frac{1}{n}\right)^n < 1$, we have $$\left(1+\frac{1}{n}\right)^n+\left(1-\frac{1}{n}\right)^n < 4 < n +\frac{1}{n}$$ Lemma 2 For any $k$ and $n \geq 4$, $$n^{n+k} -(n+k)^n > n^{n+1} - (n+1)^n$$ Proof \begin{align*} &\qquad n^{n+k} -(n+k)^n > n^{n+1} - (n+1)^n \\ &\Leftrightarrow n^{n+k} -n^{n+1} > (n+k)^n - (n+1)^n \\ &\Leftrightarrow n^k - n > \left(1+\frac{k}{n}\right)^n - \left(1+\frac{1}{n}\right)^n \end{align*} Since $\left(1+\frac{k}{n}\right)^n < 3^k$ and $\left(1-\frac{1}{n}\right)^n < 1$, we have $$\left(1+\frac{k}{n}\right)^n+\left(1-\frac{1}{n}\right)^n < 3^k +1 < n^k - n $$ as $n > 4$.

Now suppose that $a^b - b^a = 3$. Let $a \geq 4$. If $b = a+k > a$, then we have \begin{align*} a^{a+k} - (a+k)^a &> a^{a+1} - (a+1)^a \\ &> (a-1)^a - a^{a-1} \\ &> \cdots \\ &> 3^4 - 4^3 = 17 \end{align*} Thus there are no solutions with $b > a \geq 4$.

If $a > b \geq 4$, then $$a^b - b^a = -(b^a - a^b) \leq -17 $$ from what we have seen above. Thus there are no solutions if $a > b \geq 4$. Thus all solutions can be only in the range $1 \leq a,b \leq 4$. Clearly, in this range only $a=4, b=1$ satisfies the given equation.

WolframAlpha agrees with your answer. Afraid I can't help tell with answering your question though, sorry — lioness99a, Mar 20 '17 at 14:55
Compare also eith this MSE-question; it seems to be a classical question, also for $x^y-y^x=k$. — Dietrich Burde, Mar 20 '17 at 15:00
@DietrichBurde Thanks for the comment. The first reference is helpful. It appears possible that a bound can be worked out to conclude that $a,b$ should be less than 5. I will try and post a solution (if I get one!!!) — , Mar 20 '17 at 15:26
According to the table here : https://en.wikipedia.org/wiki/Catalan%27s_conjecture , the only pairs of perfect powers with difference $3$ are $(1,4)$ and $(125,128)$, but the second pair is not of the form $a^b,b^a$. But as already pointed out, an easier proof probably exists. — Peter, Mar 20 '17 at 15:52

score 5 · Answer 1 · answered Mar 20 '17 at 18:29

Consider $$f_a(x)=x\ln(a)-a\ln(x)$$ with $a,x\ge 3$

We have $$f_a'(x)=\ln(a)-\frac{a}{x}>0$$ for $x>\frac{a}{\ln(a)}$

In particular, $f_a(x)$ is strictly increasing for $x\ge a$ and we have $$f_a(a+1)=(a+1)\ln(a)-a\ln(a+1)$$

The function $g(x)=(x+1)\ln(x)-x\ln(x+1)$ has derivate $g'(x)=\ln(x)-\ln(x+1)+\frac{x+1}{x}-\frac{x}{x+1}=\ln(\frac{x}{x+1})+\frac{2x+1}{x^2+x}>\ln(\frac{x}{x+1})+\frac{1}{x}=\frac{1}{x}-\ln(1+\frac{1}{x})>0$ for $x\ge 3$

So, we have $f_a(x)\ge 0.235566$ for $3\le a\le x$ hence $$x\ln(a)\ge a\ln(x)+0.235566$$ implying $$a^x>1.26x^a$$

Since $x^a\ge 27$, we can conclude $$a^x>x^a+7$$

This rules out $3\le a<x$

The case $3\le x<a$ can be ruled out analogue by changing the roles of $a$ and $x$. $a=x$ can be rules out anyway.

Now, we show $2^a>a^2+3$ for $a\ge 5$ by induction

$a=5$ : $2^5=32>28=5^2+3$

$$2^{a+1}=2\cdot 2^a>2(a^2+3)=2a^2+6>a^2+2a+4=(a+1)^2+3$$

for $a^2>2a-2$ , or $a^2-2a+2=(a-1)^2+1>0$ , which is true for every $a$

The case $x=1$ and the remaining cases for $x=2$ can be verified easily.

Andre.J · Accepted Answer · 2017-03-20T20:04:13.713

Consider $g(t) = a^t - t^a .$ Then $g'(t) = (\ln a) a^t - at^{a-1}$.

Note that $$ \left(\frac{t}{\ln (t)} \right )' = \frac{(-1 + \ln(t))}{\ln^2(t)}$$ and $$\left( \frac{t-1}{\ln(t)}\right)' = \frac{t(\ln (t) -1)+1}{t \ln^2(t)} $$ Thus both $t /\ln t$ and $(t-1)/\ln t$ are strictly increasing for $t > e$. Therefore, for $a,t > e$ and $t >a$, $a^t > t^a$ and $$ \frac{t-1}{\ln t} > \frac{a-1}{\ln a}$$ $$\Rightarrow a^{t-1} > t^{a-1} $$ $$\Rightarrow a^t > at^{a-1} $$ $$\Rightarrow (\ln a)a^t > at^{a-1}$$ That is, for $a,t > e$ and $t >a$, $g(t)$ is strictly increasing. Thus for positive integers $ n, m \geq 3, n \neq m$, $|n^m-m^n| = n^m - m^n $ if $m >n$ and $m^n - n^m$ if $n > m$. Since g is increasing, we have $|n^m-m^n| \geq |n^{n+1}-(n+1)^n|$ or $|n^m-m^n| \geq |m^{m+1}-(m+1)^m|$.

Now, note that $h(x) = x^{x+1} - (x+1)^x$ is increasing for $x \geq 3$. Thus $|n^m-m^n| > |3^4-4^3| = 17 > 3$. Thus no positive integer solutions exist for $n,m \geq 3$. The cases $n$ or $m \in \{1,2\}$ are easily verified, and the solution you found is the only one.

Find all solutions in positive integers to $a^b -b^a = 3$

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