I think I can prove this mathematically without using binomial theorem or any higher level math. Let mw know what you think :Mathematical Proof\(a^b=b^a\)
without loss of generality, I assume a>b. therefore, \(a=b^{a/b}\)
Also it is easy to see any solution (a,b) has to be either both positive integers or both negative (Otherwise one side will become fractional with abs value less than 1, and the other side will have an abs value>1)
Assume for the moment, a,b>0
Since a and b are integers, a>b, (a/b)>1. For \(b^{a/b}\) to be an integer, a/b must be an integer. Thus in all cases, a has to be a multiple of b whenever this equation has a solution. Let a=kb, where k is an integer, such that k cannot be 1, since in that case a=b which is not allowed
Hence, \(kb=b^k\)
\(b^k-bk=0\)
\(b(b^{k-1}-k)=0\)
Either b=0 OR \(b=k^{\frac{1}{k-1}}\)
b=0 implies a=kb=0. Hence it is an invalid solution.
Thus only solutions are given by \(b=k^{\frac{1}{k-1}}\)
Clearly for k=2, this is an integer \(b=2^1=2\) & \(a=bk=4\)
For any higher k, this is not an integer :
k=3, b=3^(1/2)
k=4, b=4^(1/3)
k=5, b=5^(1/4)
.. and so on ... which always gives irrational values of b which are not permitted
Hence only solution (positive) is b=2,a=4
Now notice that if (a,b) is a solution then (-a,-b) is also always a solution as long as both a and b are either odd or both are even which is the case here.
Hence b=-2,a=-4 is also a solution
By symmetry the pairs b=4,a=2 & b=-4,a=-2 are also solutions
Hence overall there are four pairs of solutions. And we have proved there can be no more feasible solutions
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