Commutativity between Base and Exponent

• Note that if $(p,q)$ is a solution, then $(q,p)$ is also a solution to the equation $a^b=b^a$ .

• None of $a$ and $b$ can be $1$ .

• Without the loss of generality, let $a<b$ .

• $a \mid b$ . Otherwise, $a^b\equiv 0 \not \equiv b^a (\mod a$ ).

• Then $ak=b$ , for some positive integer $k\geq2$ .

• Now, $a^b=b^a \iff a^{ak}=b^a \iff (a^k)^a=b^a \iff a^k=b$ (As none of $a^k$ and $b$ is $1$ or $0$ , and $a\neq 0$ ).

• $ak=b$ and $a^k=b \implies ak=a^k \implies k=a^{k-1}$ (As $a\neq 0$ ).

• So, $\sqrt[k-1]{k}=a$

• For $k=2$ , $a=\sqrt[k-1]{k}=2$ and $b=ak=4$ . $(2,4)$ and $(4,2)$ are solutions to the equation.

• It can be proved by Induction on $k$ that

  For $k\geq 3$ , $a^{k-1}>k$ , with $a\geq2$ ;

which implies $a^{k-1} \neq k \implies$ there are no more solution $(a,b)$ , other than $(2,4)$ and $(4,2)$ , to the equation $a^b=b^a$ .

So, $(2+4)+(4+2)=\boxed{12}$ is the answer.

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Let's prove that for $k\geq 3$ , $a^{k-1}>k$ , with $a\geq2$ .

Induction Basis (IB) : For $k=3$ , $a^{k-1}=a^2\geq2^2=4>k=3$ . So, for $k=3$ , the claim holds.

Induction Hypothesis (IH) : Let's assume for some $m\geq 3$ , $a^{m-1}>m$ .

Inductive Step(IS) : $a^{m-1}>m \implies a^{m-1}\times a > am\geq2m>m+1$ (As $m\geq3>1$ ) $\implies a^m>m+1$ . So, the claim holds for $k=m+1$ provided it holds for $k=m$ .

Hence, By the Principle of Mathematical Induction, for $k\geq 3$ , $a^{k-1}>k$ , with $a\geq2$ . (Proved)