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> Given two integers $a \ge b >1$, can we encode them as a unique integer $a^b + b^a$? 
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I [asked](https://math.stackexchange.com/questions/888009/for-integers-a-ge-b-ge-2-is-fa-b-ab-ba-injective) this question on math.SE, and after surviving a week with a bounty, it seems that this question is harder than I initially thought.


Apparently these things have been named [Leyland Numbers](http://en.wikipedia.org/wiki/Leyland_number), but none of the literature I've been able to find on them provides proof that there are no repeats.



For integers $a \ge b > 1$ is $f(a,b) = a^b + b^a$ injective?
I will show that if we assume the $abcd$ conjecture (which is the case $n=4$ of Browkin and Brzezinski's [$n$-conjecture](http://cr.yp.to/bib/1994/browkin.pdf) that generalizes the $abc$ conjecture), then $a^b+b^a=c^d+d^c$ has only finitely many solutions with $\{a,b\}\neq \{c,d\}$.


The $abcd$ conjecture claims that if $a,b,c,d$ are integers with $a+b+c+d=0$ and nonzero subsums (i.e. $a+b+c\neq 0$ etc.), $\gcd(a,b,c,d)=1$ and $\varepsilon>0$, then 
$$\max\{|a|,|b|,|c|,|d|\}\leq K\_{\varepsilon} \text{rad}(abcd)^{3+\varepsilon}$$
(it is enough for us to assume the conjecture with any absolute constant in place of $3+\varepsilon$). Here $\text{rad}(m)$ is the product of the prime divisors of $m$. Given that Mochizuki seems to have proved the $abc$ conjecture and some generalizations of it, perhaps the $abcd$ conjecture is not that distant an assumption. 


If $a^b+b^a=c^d+d^c$ and $\gcd(a,b,c,d)=1$, we obtain
$$\max\{a^b,b^a,c^d,d^c\}\leq K\_{\varepsilon}\text{rad}(a^bb^ac^dd^c)^{3+\varepsilon}=K\_{\varepsilon}\text{rad}(abcd)^{3+\varepsilon}\leq K\_{\varepsilon}(abcd)^{3+\varepsilon},$$ unless some subsum of $a^b+b^a-c^d-d^c=0$ is zero, which would give $a^b=c^d$ and $b^a=d^c$ (or vice versa), but then $a$ and $c$ have the same prime factors, and writing $a=\prod\_{i=1}^s p\_i^{\alpha\_i}, b=\prod\_{i=1}^s p\_i^{\beta\_i}$, we see from $a^b=c^d$ that $\frac{\alpha\_i}{\beta\_i}=\frac{d}{b}$, which is independent of $i$, so $a\mid c$ or $c\mid a$. Similarly $b\mid d$ or $d\mid b$. After this it is easy to see that $a^b=c^d$, $b^a=d^c$ has no nontrivial solutions. In fact, if for instance $d=kb,a=\ell c$, then after simplification the equations become $\ell=c^{k-1},k=b^{\ell-1}$, so $k\geq 2^{\ell-1}$ and then $\ell\geq 2^{2^{\ell-1}-1}$. Hence $\ell=2$ and similarly $k=2$ (or $\{a,b\}=\{c,d\}$), but then $b=c$, and thus $a=d$.


Now we may assume that the subsums are nonzero. Choose $\varepsilon=1,$ say, and let $d=\max\{a,b,c,d\}$. Then $2^d\leq c^d\leq K\_1\text{rad}(abcd)^4\leq K\_1\cdot d^{16}$, so $d$ is bounded by an absolute constant, and hence $a,b,c,d$ are all bounded.


Now assume $r=\gcd(a,b,c,d)>1$. The next step is to show that $r$ is bounded. Now $a^b+b^a=c^d+d^a$ is of the form $x^r+y^r=z^r+w^r$, where $x=a^{\frac{b}{r}},...,w=d^{\frac{c}{r}}$. We will show that if $r$ is large and $(x,y,z,w)$ is any quadruple of positive integers satisfying $x^r+y^r=z^r+w^r$, then $\{x,y\}=\{z,w\}$. By homogeneity, it suffices to show that the coprime solutions satisfy $\{x,y\}=\{z,w\}.$ Since we were allowed to make the assumption $\gcd(x,y,z,w)=1$, the $abcd$ conjecture implies 
$$\max\{x^r,y^r,z^r,w^r\}\leq K\_1\text{rad}(x^ry^rz^rw^r)^4\leq K\_1(xyzw)^4,$$
unless a subsum of $x^r+y^r-z^r-w^r$ vanishes, which leads to $\{x,y\}=\{z,w\}$.
If the subsums are nonzero and $w=\max\{x,y,z,w\}$, then $w^r\leq K\_1w^{16}$, so $r$ is bounded or $w=1$. The last case leads to $x=y=z=w=1$. Therefore, for large $r$, the only solutions to $x^r+y^r=z^r+w^r$ are those where $\{x,y\}=\{z,w\}$. Thus also $\{a^b,b^a\}=\{c^d,d^c\}$, which was already seen to give no nontrivial solutions.


Finally, let $M$ be an absolute constant that is an upper bound for $r$. Let $R=\gcd(a^b,b^c,c^d,d^c)$. We apply the $abcd$ conjecture once again to see that $\max\{\frac{a^b}{R},\frac{b^a}{R},\frac{c^d}{R},\frac{d^c}{R}\}\leq K\_1\text{rad}(abcd)^4$. Now if $abcd$ has no prime divisor greater than $M$, we have 
$$\min\{a^b,b^a,c^d,d^c\}\geq R\geq c\_0\max\{a^b,b^a,c^d,d^c\}$$
for some absolute constant $c\_0>0$. Now if there exists a prime $p\_1$ that divides some of $a,b,c,d$ but not all of them, then
$$R\leq \prod\_{p\leq M, p\neq p\_1}p^{\min\{v\_p(a^b),v\_p(b^a),v\_p(c^d),v\_p(d^c)\}}\leq \frac{\max\{a^b,b^a, c^d, d^c\}}{2^{\min\{a,b,c,d\}}}\quad \quad (1).$$
If no such $p\_1$ exists, all the numbers $a,b,c,d$ have the same prime factors. Let $a=\prod\_{i=1}^s p\_i^{\alpha\_i},...,d=\prod\_{i=1}^s p\_i^{\delta\_i}$. The condition $\gcd(a,b,c,d)\leq M$ tells $\min\{\alpha\_i,\beta\_i,\gamma\_i,\delta\_i\}\leq M$. Let $P^{\delta}$ be the largest prime power dividing $d$. Then $\min\{v\_{P}(a),v\_{P}(b),v\_{P}(c)\}\leq M$ if $d$ is large. For example, let $v\_{P}(c)\leq M$. Write $d=P^{\delta}D,c=P^{\gamma}C$, where $P\nmid C,D$. Then 
$$R=\gcd(a^b,b^a,c^d,d^c)\leq \gcd(c^d,d^c)\leq P^{Md} D^c=P^{Md-c\delta}d^c\leq M^{Md}d^{-\frac{c}{s}}d^c,$$
where $s\leq M$ is the number of prime factors of $d$. The last quantity is at most $\left(\frac{d}{2}\right)^c$ if $\frac{1}{2}d^{\frac{1}{s}}\geq M^{\frac{Md}{c}},$ which holds for large enough $d$ if $\frac{d}{c}$ is bounded. It must indeed be bounded since we had $\frac{c^d}{d^c}\geq c\_0$. Therefore $(1)$ holds again for all large $d$. 


Next we show that $(1)$ holds also if $a,b,c,d$ have some prime factors greater than $M$, and then use $(1)$ to deduce a contradiction.


Now one of $a,b,c,d$ has a prime factor $p\_0>M$. For example, let $p\_0\mid d$. Then 
$$R\leq \prod\_{p\leq M}p^{v\_p(d^c)}\leq \left(\frac{d}{p\_0}\right)^c\leq \frac{\max\{a^b,b^a,c^d,d^c\}}{2^{\min\{a,b,c,d\}}},$$
so $(1)$ must always hold for large $d$. However, we had $R\geq \frac{\max\{a^b,b^a,c^d,d^c\}}{K\_1\text{rad}(abcd)^4}\geq \frac{\max\{a^b,b^a,c^d,d^c\}}{K\_1 d^{16}}$. Thus $K\_1d^{16}\geq 2^{\min\{a,b,c,d\}}$. But  

$$\min\{a,b,c,d\}^d\geq \min\{a^b,b^a,c^d,d^c\}\geq R\geq \frac{\max\{a^b,b^a,c^d,d^c\}}{K\_1d^{16}}\geq \frac{c^d}{K\_1d^{16}},$$
implies $\min\{a,b,c,d\}\geq k\_0c$ for some constant $k\_0$,so $K\_1d^{16}\geq 2^{k\_0c}$. Still we have $\frac{d^c}{c^d}\geq \frac{1}{K\_1d^{16}}$. In particular, $d^c\geq \frac{2^d}{K\_1d^{16}}$, so $c\geq k\_2 \frac{d}{\log d}$. But then $K\_1d^{16}\geq 2^{k\_0c}\geq 2^{k\_3\frac{d}{\log d}}$ shows that $d$ is bounded and hence all $a,b,c,d$ are bounded. 


