Where do I start?
find all the integer solutions (a,b) that satisfy the conditions:
a^b = b^a,
a and b do not equal zero,
a does not equal b.
Give your answer as the sum of all the solutions; for example, if the solutions were (a,b) = (6,7) or (7,6) then the answer would be 26.
The answer is 12.
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1 solution
Let's assume that b and a are co-prime, then a^b cannot equal b^a as each side share no common factors.
Lets assume that b = k x a^c where k and c are unknown constants, k >= 1 and k and a are coprime. then as b^a =a^b (k x a^c)^a = a^(k x a^c) k^a x a^ac = a^(k x a^c) There is only one prime factor on the right hand side and at least two prime factors on the left hand side. As k and a are coprime these two sides cannot be equal hence there is a contradiction.
Therefore; b = a^k where k is an integer.
as a^b = b^a
a^(a^k) = (a^k)^a
a^(a^k) = a^(ak)
a^k = ak
a^(k-1) = k
If k = 0, b = 1 then a = 1 which doesn't satisfy the conditions so it can be disregarded.
If k = 1 the b = a^1 = a which doesn't satisfy. the conditions.
If k = 2 a^2 = 2a a = 2 b = 4 The solution works equally well when b = 4 and a = 2.
If k >= 3 Then we can prove that the equation 2^(k-1) > k for all values of k >= 3 by induction.
Step 1: Show for k = 3:
LHS = 2^(3-1) = 4
RHS = 3
LHS > RHS
Step 2:
Assume for k = c
2^(c-1) > c
Step 3:
Prove for k = c + 1
LHS = 2^(c) = 2 x 2^(c-1) > 2c > c+1 we can assume this because c >= 3
RHS = c+1
Therefore; LHS > RHS.
Therefore; 2^(k-1) > k for values of k >= 3.
This means that 2^(k-1) = k has no integer solutions for k >= 3. From this we can deduce that a^(k-1) = k also has no solutions if a >= 2. So the solutions for (a,b) are (2,4) and (4,2).
The sum of all these solutions is 12.
This is fairly similar to a solution I made last year but it was overly complicated so a hat-tip to Dr. Math for formulating it a bit more clearly and just so I could check I was right. I also found another solution but it is very nasty so anybody who has a better proof please post it.