Suppose *a* and *b* are different and greater than 1 (otherwise
the problem is trivial).
Then *a* can be written *a* = *b*^{x}
where *x* is a real number. If this is inserted into the equation
*a*^{b} = *b*^{a}
we get (*b*^{x})^{b}
= *b*^{a} or *b*^{xb}
= *b*^{a}, which gives *xb*
= *a*. But then *x* is rational, so *x* = *p*/*q*
where *p* and *q* are relatively prime positive integers.

From this we get *a*^{q} = *b*^{p}.
Now, write *a* and *b* as products of primes. From the uniqueness
of factorization it follows that any prime in the *a*-product must
occur raised to an exponent which is a multiple of *p*. Thus,
*a* = *c*^{p} for some integer
*c *> 1. Inserting this into *a*^{q}
= *b*^{p} we get *b*
= *c*^{q}. If then the expressions for
*a* and *b* are inserted into *a*^{b}
= *b*^{a} we get *pc*^{q}
= *qc*^{p} or *c*^{p-q}
= *p*/*q*.

If *p* > *q* then *q* = 1, since *p*
and *q* are relatively prime, so *c*^{p-}^{1}
= *p*. This is satisfied by (*c*,*p*) = (2,2), which
gives (*a*,*b*) = (4,2) as one solution of the problem.
Now 2^{p-1} > *p *
if *p* > 2, and *c*^{p-}^{1}
> 2^{p-1} if *c* >
2, *p* > 1, so there are no more possibilities with *p*
> *q.* Similarly *q* > *p* gives (*a*,*b*)
= (2,4).

Thus, (*a*,*b*) = (2,4), (*a*,*b*) =
(4,2) or *a* = *b*.

**Note:** The inequality 2^{p-1}
> *p* can be proved by induction, and *c*^{p-}^{1}
> 2^{p-1} follows from a factorization
of *c*^{p-}^{1}
- 2^{p-1} into the factor
*c* - 2 and one positive factor.

There is quite another method for the equation *a*^{b}
= *b*^{a}. Like above we suppose *a*
and *b* are different and greater than 1.
Define *f*(*x*) = (ln *x*)/*x* for
*x* > 0. Then the equation *a*^{b}
= *b*^{a} is equivalent to
*f*(*a*) = *f*(*b*). The function *f* is strictly
increasing for 0 < *x* < *e*, and strictly decreasing
for *x* > *e*. Thus, if *f*(*a*) = *f*(*b*)
for 0 < *a* < *b*, then 0 < *a* <
*e*. Now, the only integer between 1 and *e* is 2. So for a non-trivial
solution of *a*^{b} = *b*^{a}
with *a* < *b* we must have *a* = 2.
This implies *b* = 4, since *f* is decreasing for
*x* > *e*. Then, by symmetry the non-trivial solutions are (2,4)
and (4,2).

Can this method by used for the IMO-problem? I don't know so tell me
if you succed, please!