Find integral solution of $a^{b} - b^{a} = 3$ Find the integral solution of $$a^b -  b^a = 3$$
I am a student of class 10 and got this question in a maths competetive exam.
I have tried converting into a log equation in one variable, or solve it by using parity but couldn't solve it.
Thanks in advance

I suppose you are familiar with the inequality $\ln(1+x)\le x$. Now if $b>a$, then $$b^a=a^a\left(1+\frac{b-a} a\right)^a=a^a e^{a\ln(1+(b-a)/a)}\le a^a e^{b-a}.$$
When $a\ge 3$, we have
$$3=a^b-b^a\ge a^a(a^{b-a}-e^{b-a})\ge a^a(a-e)\ge27(3-e)>7$$
a contradiction. So $a\le2$ if $b>a$. But it can be checked that there is no solution for $a=1$ (obvious) and $a=2$ (since $2^b-b^2>3$ for $b\ge 5$). We conclude that there is no solution satisfying $b>a$. But $b\neq a$, so the only possibility is that $b<a$.  
However, for $3\le b<a$, it's well-known that $a^b<b^a$. So it's only possible when $b\le 2$. For $b=1$ the solution is $(a,b)=(4,1)$. For $b=2$, due to the same reason above($a^2-2^a<-3$ for $a\ge 5$), there is no solution.  
To conclude, the only solution is $(a,b)=(4,1)$.
