# Find all positive integers satisfying $a^{b^2}=b^a$.

So the question is :

Determine all pairs $$(a, b)$$ of positive integers satisfying $$a^{b^2} = b^a$$.

I tried it for 2 hours by different methods like taking it even, odd and by modulus method but cannot find any solution. Please help me in this question.

• Running a quick Mathematica program, I found that if $b\le1000$ then $(b,a)\in\{(1,1),(2,16),(3,27)\}$ are the only possible solutions. – Vepir Jan 10 '20 at 16:10

Just follow your nose; let $$d:=\gcd(a,b)$$ so that $$a=du$$ and $$b=dv$$ with $$u$$ and $$v$$ coprime. Then $$b^a=(dv)^{du}=((dv)^u)^d \qquad\text{ and }\qquad a^{b^2}=(du)^{d^2v^2}=((du)^{dv^2})^d,$$ from which it follows that $$(dv)^u=(du)^{dv^2}$$. Because $$u$$ and $$v$$ are coprime we either have $$u=1$$ or $$v=1$$.
If $$u=1$$ then $$dv=d^{dv^2}$$, and so $$v=d^{dv^2-1}$$ from which it quickly follows that also $$v=d=1$$ and hence $$a=b=1$$.
If $$v=1$$ then $$d^u=(du)^d$$ from which it follows that $$u^d=d^{u-d}$$, and in particular $$u\geq d$$. Let $$c:=\gcd(d,u)$$ so that $$d=ce$$ and $$u=cw$$ with $$e$$ and $$w$$ coprime and $$w\geq e$$. Then $$u^d=(cw)^{ce}=((cw)^e)^c \qquad\text{ and }\qquad d^{u-d}=(ce)^{cw-ce}=((ce)^{w-e})^c,$$ from which it follows that $$(cw)^e=(ce)^{w-e}$$. As $$e$$ and $$w$$ are coprime and $$w\geq e$$ it follows that $$e=1$$, so $$cw=c^{w-1},$$ and hence $$w=c^{w-2}$$, from which it quickly follows that $$w\leq4$$. We check these few cases:
1. If $$w=1$$ then $$c=d=1$$ which leads to $$a=b=1$$.
2. If $$w=2$$ then $$u=2d$$ and hence $$a=2b^2$$, and plugging this in shows that $$b^{2b^2}=(2b^2)^{b^2}=2^{b^2}b^{2b^2},$$ contradicting the fact that $$b$$ is a positive integer.
3. If $$w=3$$ then $$c=d=3$$ which leads to $$a=27$$ and $$b=3$$.
4. If $$w=4$$ then $$c=d=2$$ which leads to $$a=16$$ and $$b=2$$.