This is my answer for the following question: Find all natural numbers $(a,b)$ for which $a^b-b^a=1$.
When $a$ or $b$ equals $1$, $(a,b)=(2,1)$ is trivial. If $a,b>1$, I generalized the problem to $b^a|a^b-1$
Let $p$ be the smallest prime divisor of $b$. Then $a^ {p-1} \equiv a^ {b} \equiv 1\pmod p$.
Since $p$ is the smallest prime divisor of $b$, $gcd(b,p-1)=1$. Therefore $gcd(a^b-1,a^ {p-1}-1)=a-1 \equiv 0\pmod p$. $\therefore a \equiv 1\pmod p$.
By L.T.E
If $p$ is not $2$, $a{ v }_{ p }(b)\le { v }_{ p }(a^{ b }-1)={ v }_{ p }(a-1)+v_{ p }(b)$. Thus, ${ v }_{ p }(a-1)\ge a-1$. A contradiction.
If $p=2$,
$I)$ $a \equiv 1\pmod 4$: This implies that $a{ v }_{ 2 }(b)\le v_{ 2 }(a-1)+v_{ 2 }(b)$. Thus ${ v }_{ 2 }(a-1)\ge a-1$. A contradiction.
$II)$ $a \equiv 3\pmod 4$: This implies that $a{ v }_{ 2 }(b)\le v_{ 2 }(a+1)+v_{ 2 }(b)$. Thus ${ v }_{ 2 }(a-1)\ge a+1$, implying $a=3$ and ${v}_{2}(b)=1$.
If $b=2B, B \equiv 1\pmod 2$, $B^3|3^ {2B}-1$. If the smallest prime divisor of $B$ is $q$, $q$ is not $2$. However, $q|3^ {2B}-1$ and $q|3^ {q-1}-1$. Therfore $q|3^2-1$, a contradiction. Therfore $B=1$.
Thus the only solution is $(3,2),(2,1)$.
However, this solution seemed a bit strange. I think that there must be other solutions to this problem without using Catalan`s Conjecture. What are other elementary ways to solve this equation?
