Inspired by Satyajit Mohanty

All variables introduced in this solution are positive integers.

It is obvious that $b>a$ .

Write $a=kc, b=kd$ where $k=\gcd(a,b) \geq 1$ . Substituting into the equation yields:

${(kc)}^{k^{5}c^{5}}={(kd)}^{kd}$

${k}^{k^{5}c^{5}-kd}=\frac{d^{kd}}{c^{k^{5}c^{5}}}$

Since $\gcd(c,d)=1$ and the L.H.S. is an positive integer, this is only possible if $c=1$ .

Therefore, we can write $b=ad, d \in \mathbb{N^{+}}$ . Substituting this into the original equation, we have

$a^{a^{5}}=d^{da}a^{da}$

$a^{a^{5}-da}=d^{da}$

$a^{a^{4}-d}=d^{d}$

Note that if $a>d$ , then L.H.S. > R.H.S. as long as $a>2^{\frac{1}{3}}$ since $a^{4}>2a>2d$ .

Therefore, $a\leq d$ and we can write $d=ea, e \in \mathbb{N^{+}}$ by the same argument above. Repeating the procedure above, we get

$a^{a^{3}-2e}=e^{e}$ .

Note that if $a>e$ , then L.H.S. > R.H.S. as long as $a>\sqrt{3}$ since $a^{3}>3a>3e$ .

Therefore, $a \leq e$ and we can write $e=fa, f \in \mathbb{N^{+}}$ . Repeating the procedure above, we get

$a^{a^{2}-3f}=f^{f}$

Note that if $a > f$ , then L.H.S. > R.H.S. as long as $a>4$ since $a^{2}>4a>4f$ .

We will manually check the cases for $a=2,3,4$ before proceeding to assume that $a \leq f$

If $a=2$ , note that $2^{4-3f}=f^{f}$ . Since $a>f$ , this forces $f=1$ , but $2^{4-3} \neq 1^{1}$ .

If $a=3$ , note that $3^{9-3f}=f^{f}$ . Since $a>f$ , this forces $f=1$ or $2$ , but the equation does not hold in either cases.

If $a=4$ , note that $4^{16-3f}=f^{f}$ . Since $a>f$ , this forces $f=1,2$ or $3$ , but equation does not hold in all $3$ cases.

Therefore, with $a \leq f$ , we can write $f=ga, g \in \mathbb{N^{+}}$ . Repeating the procedure above, we get

$a^{a-4g}=g^{g}$ .

If we assume $a \leq g$ and write $g=ha, h \in \mathbb{N^{+}}$ , we get

$a^{1-5h}=h^{h}$ which is impossible since we require $1-5h>0$ if $h$ is to be a positive integer. Therefore, we can conclude that $a>g$ .

Now, write $a=jg$ with $j>1, j \in \mathbb{N^{+}}$ . We get

$(jg)^{jg-4g}=g^{g}$

$(jg)^{j-4}=g$

$j^{j-4}=g^{5-j}$

The only possible value $j$ can take is $4$ . (You may prove this as an exercise.)

Solving the equation, we find that $g=1$ . Therefore, $a=4$

Write $4^{4^{5}}=4^{1024}=4^{(4)(256)}=256^{256}$ and we are done.

The answer is therefore $4+256=\boxed{260}$

Let $b=a^x×k$ where $x$ is such that $a$ does not divide $k$ . After plugging this the equation becomes $k^k=a^{a^{5-x}-x}$ . Since $a$ does not divide $k$ there must exist a prime divisor $p$ of $a$ and $k$ such that $v_{p}(a)>v_{p}(k)$ . The power of $p$ must be same in $RHS$ and $LHS$ thus giving $k=\frac {(a^{5-x}-x)×v_{p}(a)}{v_{p}(k)}>a^{5-x}-x$ --- $(i)$ . As $b<a^5$ we have $0 \leq x \leq 4$ . Now we have to just check all cases from $x=0$ to $x=4$ . Cases $x=0,1,2$ are easy to prove so I will show only $x=3,4$ .
Case $x=3$ . From $b<a^5$ and $(i)$ we get $a^2-2 \leq k <a^2$ . $k$ cannot be $a^2-1$ as it $gcd(a+1,a)=1$ . So $k$ is $a^2-2$ but this is not correct either as $LHS$ becomes greater than $RHS$ . No solutions in this case.

Case $x=4$ From $b<a^5$ and $(i)$ we get $a-3 \leq k <a$ . Again we have to check $k=(a-3),(a-2),(a-1)$ which is not very dufficult. After checking all cases in this case we get only one solution $a=4,k=1$ i.e. $a=4,b=256$ .