|Find all pairs (a,b) of positive integers such that a(b^2) = ba .|
|Find all pairs (a,b) of positive integers such that ab = ba .|
Then a can be written a = bx where x is a real number. If this is inserted into the equation ab = ba we get (bx)b = ba or bxb = ba, 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 aq = bp. 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 = cp for some integer c > 1. Inserting this into aq = bp we get b = cq. If then the expressions for a and b are inserted into ab = ba we get pcq = qcp or cp-q = p/q.
If p > q then q = 1, since p and q are relatively prime, so cp-1 = p. This is satisfied by (c,p) = (2,2), which gives (a,b) = (4,2) as one solution of the problem. Now 2p-1 > p if p > 2, and cp-1 > 2p-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 2p-1 > p can be proved by induction, and cp-1 > 2p-1 follows from a factorization of cp-1 - 2p-1 into the factor c - 2 and one positive factor.
(a,b) = (1,1), (a,b) = (16,2) or (a,b) = (27,3).
Define f(x) = (ln x)/x for x > 0. Then the equation ab = ba 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 ab = ba 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!