Find solution for $|p^q-q^p|\leq 2$

Question

Here's a question that I saw:

Solve the inequality $|p^q-q^p|\leq 2$, where $p,q$ are primes.

Here is my attempt (although there's not much to it):

Consider $p=3,q=2$ or $p=2,q=3$. By Mihăilescu's theorem, this is the only possible solution for $|p^q-q^p|=1$.

But then what should I do next? I have no idea. Please help.

fyi, equations like $p^n-q^m=1$ may also be solved using more elementary tools such as Zsigmondy's Theorem or in some cases the Lifting The Exponent Lemma — Bart Michels, Apr 06 '17 at 13:54
You can prove that if $p+1<q$ and $p,q$ are primes, then $p^q>q^p+2$. — didgogns, Apr 06 '17 at 13:57

score 5 · Accepted Answer · edited Apr 13 '17 at 12:21

I will prove that for any positive integers $a,b$ which satisfies $2<a<b$, $a^b>b^a+2$.

Indeed, $3^4>4^3+2$ and if $3<a$, then $a>e+1>\left(1+\frac{1}{a}\right)^a+\frac{2}{a^a}$ therefore $a^{a+1}>(a+1)^a+2$ for $a>2$. Also, $$a^{k+1}>a^k\cdot a>a(k^a+2)>ek^a+2>\left(1+\frac{1}{k}\right)^ak^a+2=(k+1)^a+2$$for $k\ge a$ we can use mathematical induction.

Therefore, at least one of $p,q$ is $2$ unless $p=q$. WLOG, let $p=2$ then $q<5$, if not, $2^q>q^2$.

Now one can get all the solutions of the inequality.

Find solution for $|p^q-q^p|\leq 2$

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