Let $a-b=k\in \mathbb{Z}^+$. We get that $k^{b(b+k)}=(b+k)^bb^{b+k}$. Let $d=\gcd (b,k)$ and $c,\ell\in \mathbb{Z}^+$ such that $b=dc,k=d\ell$ where $\gcd (c,\ell )=1$, we get $d^{b(b+k)}\ell^{b(b+k)}=d^b(c+\ell )^bd^{b+k}c^{b+k}$. So $d^{b(b+k)-b-(b+k)}l^{b(b+k)}=c^{b+k}(c+\ell )^b$ but since $\gcd (\ell ,c)=\gcd (\ell ,c+\ell )=1$ so $\ell=1$. So $d^{dc(dc+d)-dc-(dc+d)}=c^{dc+d}(c+1)^{dc}\implies d^{dc^2+dc-c-(c+1)}=c^{c+1}(c+1)^c$. If $c>1$. For any prime $p$ such that $p\mid c$, we get that $p\nmid c+1$, so $$\nu_p(RHS)=(c+1)\nu_p(c)=(dc^2+dc-c-(c+1))\nu_p(d).$$Since $\gcd (c+1,dc^2+dc-c-(c+1))=1$, we get $$dc^2+dc-c-(c+1) \mid \nu_p(c)\implies c\geqslant 2^{dc^2+dc-c-(c+1)}\geqslant dc^2+dc-c-(c+1)+1.$$So $3c\geqslant dc^2+dc\implies 3\geqslant dc+d=d(c+1) \geqslant 2d\implies d=1$ and $c=2$. This give $b=2,k=1,a=3$, not satisfying the equation. If $c=1$, we get $d^{2d-3}=2\implies d=2,b=2,k=2,a=4$. Finally we get that $(a,b)=(4,2)$ is the only solution.

The equation is equivalent to $a-b=a^{\frac{1}{a}}b^{\frac{1}{b}}$ Let $\alpha=a^{\frac{1}{a}}b^{\frac{1}{b}}$. We can easily see that $\alpha$ must be an integer Yielding $a=b^k<=>\alpha=b^{\frac{b^{k-1}+k}{b^k}}$ Therefore $b^k \mid b^{k-1}+k$ hence $b^{k-1}\mid k$ Hence $k=2$ yielding $(a,b)=(4,2)$

Gruberr69 wrote: The equation is equivalent to $a-b=a^{\frac{1}{a}}b^{\frac{1}{b}}$ Let $\alpha=a^{\frac{1}{a}}b^{\frac{1}{b}}$. We can easily see that $\alpha$ must be an integer Yielding $a=b^k<=>\alpha=b^{\frac{b^{k-1}+k}{b^k}}$ Therefore $b^k \mid b^{k-1}+k$ hence $b^{k-1}\mid k$ Hence $k=2$ yielding $(a,b)=(4,2)$ why $a$ has to be $b^k$?

Answer $(a,b)=(4,2)$ If $b=1$ we get $(a-1)^a=a$ which is easy to see that no solutions in $\mathbb{N}$ If $b>1$, it so easy notice that if $p|a$ $\longleftrightarrow$ $p|b$ where $p$ is prime $\Longrightarrow$ let $a=p_1^{x_1}... p_k^{x_k}$ and $b=p_1^{y_1}...p_k^{y_k}$, since $a>b$ $\Longrightarrow$ $\exists$ $t$ such that $x_t>y_t$ $\Longrightarrow$ let $a=p_t^{x_t}.a'$ and $b=p_t^{y_t}.b'$ $\Longrightarrow$ $v_p((a-b)^{ab})=v_p(a^b.b^a)$ $\Longrightarrow$ $v_p((a-b)^{ab})=v_p(a^b)+v_p(b^a)$ $\Longrightarrow$ $y_t.a.b=x_t.b+y_t.a$ $\Longrightarrow$ $\frac{y_t.a}{x_t}=\frac{b}{b-1}...(\star)$. On the other hand: $a=p_t^{x_t}.a'\geq 2^{x_t}.\geq 2x_t$ (It's so easy by induction) $\Longrightarrow$ $\frac{a}{x_t}\geq 2...(\star\star)$ and $\frac{b}{b-1}\leq 2...(\star\star\star)$ $\Longrightarrow$ replacing $(\star\star)$ and $(\star\star\star)$ in $(\star)$ we get $2 \leq$ $\frac{y_t.a}{x_t}$ $=$ $\frac{b}{b-1}$ $\leq$ $2$ $\Longrightarrow$ $\frac{y_t.a}{x_t}$ $=$ $\frac{b}{b-1}$ $=$ $2$ it's easy to see that the case of equality in $(\star\star)$ and $(\star\star\star)$ is when $(a,x_t)=(2,1);(4,2)$ and $b=2$, since $a>b$ $a=4$ and $b=2$ ... checking $(4-2)^{2.4}=2^4.4^2=256$.

It is obvious that $a\equiv b\equiv 0\pmod 2$. Case 1. $v_2(a)> v_2(b)$. $\implies v_2(a-b)=\min\left \{ v_2(a),v_2(b) \right \}=v_2(b)\implies v_2((a-b)^{ab})=ab.v_2(b)$ $\implies ab.v_2(b)=v_2(a^bb^a)=v_2(a^b)+v_2(b^a)=b.v_2(a)+a. v_2(b)$ $\implies a.v_2(b).(b-1)=b.v_2(a) \implies \frac{a}{v_2(a)}\cdot v_2(b)=\frac{b}{b-1}\implies \frac{a}{v_2(a)}\leq \frac{b}{b-1}$ $\implies \frac{2^{v_2(a)}}{v_2(a)}\leq \frac{b}{b-1}$. But it is obvious that $\frac{2^{v_2(a)}}{v_2(a)}\geq 2$ so $b$ must be $2$. $\implies a=2v_2(a)$ but $a>b=2$ so $a=4$. Case 2. $v_2(a)< v_2(b)$. Similarly the case 1 implies that $a=2$ ,contradiction. Case 3. $v_2(a)=v_2(b)=c$. Put $a=2^c.a',\ b=2^c.b'$ ($b',c'$ is odd) implies $(2^c)^{ab}(a'-b')^{ab}=2^{bc}a’^b2^{ac}b’a=(2^c)^{a+b}a'^bb'^a$ But $ab>a+b$ since $a>b;a,b$ is even ,contradiction. $\boxed{\text{Answer :} a=4,\ b=2}$

(a^1/a)*(b^1/b) is integer, easy to check the only prime factor of a and b is 2.

YadisBeles wrote: Find all ordered pairs $(a,b)$ of positive integers that satisfy $a>b$ and the equation $(a-b)^{ab}=a^bb^a$. If $a-b = 1$, then $a=b=1$ which is false. Now $a-b\geq 2$. We have \[ab \log (a-b) = b\log a + a\log b \implies \log (a-b) = \frac{\log a}{a} + \frac{\log b}{b}.\]If $b\geq e$, \[\log 2\leq \log (a-b) = \frac{\log a}{a} + \frac{\log b}{b} < 2\cdot\frac{\log b}{b}\implies b<4.\]Therefore $b = 1,2,3$. Notice that $a,b$ cannot be both odd ($a-b$ is even, but $a^bb^a$ is odd) and exactly one of $a,b$ also cannot be odd ($a-b$ is odd, but $a^bb^a$ is even). Therefore $a$ and $b$ are both odd. This gives $b = 2$ which gives $(a-2)^{2a} = a^2\cdot 2^a$. Note that if $p\mid a$ and $p\neq 2$, then $p\nmid a-2$ which is false, so $a$ and $a-2$ are powers of two. Therefore we have $(a,b) = (4,2)~~\square$

Let $a=dx,b=dy$ with $\gcd(x,y)=1$. Then $d^{d^2xy}(x-y)^{d^2xy}=d^{dx+dy}x^{dy}y^{dx}$, so $d^{d^2xy-dx-dy}(x-y)^{d^2xy}=x^{dy}y^{dx}$. Suppose a prime $p\mid x-y$. Then $p$ must divide the RHS, so it divides exactly one of $x,y$, since we cannot have $p\mid x$ and $p\mid y$. But since $p\mid x-y$, we get $p$ divides both if it divides one, contradiction. Therefore, $x-y=1$. The equation becomes \[d^{dy(y+1)-y-(y+1)}=(y+1)^{y}y^{y+1}.\]Take a prime $p\mid d$. Then \begin{align*} \nu_p(LHS) &= [dy(y+1)-2y-1]\nu_p(d),\\ \nu_p(RHS) &\le (y+1)\max\{\nu_p(y),\nu_p(y+1)\} \le d(y+1) \log_p(y+1). \end{align*}In particular, $dy(y+1)-2y-1\le d(y+1)\log_p(y+1)$ since $\nu_p(d)\ge 1$. The LHS is $O(y^2)$ while the RHS is $O(y\log y)$, so we are almost done. We have $dy(y+1)-2yd-d\le dy(y+1)-2y-1$, so \[y(y+1)-2y-1 \le (y+1) \log_p(y+1) \le (y+1)\log_2(y+1). \]Dividing by $y+1$, we get $y-\tfrac{2y+1}{y+1}\le \log_2(y+1)$, so $y-2\le \log_2(y+1)$, i.e. $2^{y-2} \le y+1$. This only holds for $y\le 4$. It is easy to check these cases, and we see that only holds for $(y,d)=(1,2)$, i.e. $(a,b)=(4,2)$.

Here are simple solutions with no explicit uses of $\nu_p$ and covering also the case $a<b$.

The solution above is too long. You can make a bound first to make things easier.

$a-b=((a-b)^{ab})^{\frac{1}{ab}}=(a^bb^a)^{\frac{1}{ab}}=a^{\frac{1}{a}}b^{\frac{1}{b}} \leq (e^{\frac{1}{e}})^2=2.08707……$ $\therefore a-b=1\vee 2$ If $a-b=1$ it is obvious that there is no solution If $a-b=2$ since $(a-b)^{ab}$ is a pure power of $2$ so both $a$ and $b$ are pure power of $2$. Combine with $a-b=2$ we know the only solution is $(4,2)$

shanelin-sigma wrote: The solution above is too long. But uses only straightforward ideas which even a well prepared 8th grader can think of Your solution is also very good, but accessible only to senior students.

Ok, here is the bounds solution in elementary terms. We have $a-b = \sqrt[a]{a}\sqrt[b]{b}$. However, $t < \left(\frac{3}{2}\right)^t$ for $t\geq 2$ by induction (base is $2 < \frac{9}{4}$ and the induction step follows by $\frac{3}{2} \geq \frac{t+1}{t}$), hence $a-b \leq \frac{9}{4}$, so $a-b$ is $1$ or $2$. If $a=b+1$, then $a^bb^a=1$, impossible. If $a-b = 2$, then $a^bb^a = 2^{ab}$, so $a$ and $b$ are powers of $2$ with difference $2$. It is not possible for both of them to be divisible by $4$ (as their difference is not), so the only possibility is $b=2$ with $a=4$.