The first we has a lemma: Consider the equation: $ a^m=b^n$ Call $ d=\gcd(m,n)$ then exist $ c\in N$ such that $ a=c^{\frac{n}{d}},b=c^{\frac{m}{d}}$ Proof We has for each $ p\in P$ then $ m{v_{p}a}=nv_{p}b$ Imply that $ \frac{n}{d}|v_{p}a$ $ \frac{m}{d}|v_{p}b$ This lemma was be prove. Now Let $ d=\gcd(a^a,b)$ then $ a^a=dm,b=dn$ where $ \gcd(m,n)=1$ Exist $ c\in N$ such that : $ a=c^n,b=c^m$ It mean that $ (c^n)^{c^n}=dm \Rightarrow c^{nc^n}=dm$ And $ dn=c^m$ So we have: $ nc^{nc^n}=mc^m$ Case 1$ nc^n\geq m$ $ nc^{nc^n-m}=m$ Imply that $ n=1$ so $ c^{c-m}=m\Rightarrow c=m=1$ So $ a=b=1$ Case 2$ m>nc^n$ so $ m=1\Rightarrow c<1$ tradition! So the equation has only solution :$ (a,b)=(1,1)$

Oh , it is the question number 4 of the second round of Iran 2014 !!!

Oh my god!

:wacko: :noo: