I've noticed that when you flip the base and the exponent in $2^4$ to get $4^2$, you get the same value, $16$. If there are any other numbers that can make this work, let me know. This is just mind-blown for me! Are there any other numbers that work? I used to do this and ta-daaaaaaah!
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3The obvious case is if the same number is used in both ;) $3^3 = 3^3$ see me swap the exponents!! – AlexR Mar 04 '15 at 21:15
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Any other tricks? – ReliableMathBoy Mar 04 '15 at 21:15
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2see http://math.stackexchange.com/questions/785836/find-all-positive-rational-solutions-to-xy-yx – Will Jagy Mar 04 '15 at 21:16
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W|A Says there are infinitely many solutions in the reals, but you probably want integers only. – AlexR Mar 04 '15 at 21:16
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Why post that right there? – ReliableMathBoy Mar 04 '15 at 21:18
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1One close vote. – ReliableMathBoy Mar 04 '15 at 21:19
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1@ReliableMathBoy Check out Martin R's duplicate link. It's a full solution. – AlexR Mar 04 '15 at 21:19
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Already six minutes into the show. – ReliableMathBoy Mar 04 '15 at 21:20
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Two close votes. – ReliableMathBoy Mar 04 '15 at 21:36
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The numbers would have to share all prime factors. In other words, if one was divisible by 7, the other would have to be as well. Only the power of each prime factor could vary.
The numbers need to be close, or the larger exponential would ruin things. Big exponentials grow fast.
2 and four work because they are close powers of 2.
A close call would be 18 and 12, but no cigars.It is close because two squared is almost equal to three. Since two to the fourth is about 17, 544 and 578 look like candidates as well. But no dice.
Assume there are two prime factors for both numbers, a and b with powers k and l for the first number and m and n for the second. Since a and b a coprime, we can ignore b and look for cases where the powers of a are equal.
The two powers of a are k times a^m times b^n and m times a^k times b^l. We can ignore b again, because a and b are coprime. We need a case where k times a^m=m times a^k. This means k and m are both powers of a. By the same token l and n are powers of b.
EDIT: Similar arguments apply to larger numbers of primes.
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In other words each of the two numbers is going to be the product of one of more stacked prime exponentials of the form a to the a to the x. That doesn't answer the question but it narrows the search. – Bernard Finucane Mar 04 '15 at 22:56