I've been playing with commutators lately, and the following puzzle turned up ... Suppose A and B are integers > 1. When is A^B / B^A an integer? What values are possible? Clearly A and B must have the same prime divisors, but the primes can occur to different powers. A = 45, B = 75 is a suggestive example. Rich
A can have prime factors that B lacks. Example: A=6, B=32. Tom
I've been playing with commutators lately, and the following puzzle turned up ...
Suppose A and B are integers > 1.
When is A^B / B^A an integer? What values are possible?
Clearly A and B must have the same prime divisors, but the primes can occur to different powers. A = 45, B = 75 is a suggestive example.
Rich
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Excellent! ------ Quoting Tom Karzes <karzes@sonic.net>:
A can have prime factors that B lacks. Example: A=6, B=32.
Tom
I've been playing with commutators lately, and the following puzzle turned up ...
Suppose A and B are integers > 1.
When is A^B / B^A an integer? What values are possible?
Clearly A and B must have the same prime divisors, but the primes can occur to different powers. A = 45, B = 75 is a suggestive example.
Rich
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In general, if you start with A and B, where A contains all of B's prime factors, then you can adjust B by scaling by A's factors. For example, if you start with A=30 and B = 20, you can keep scaling B by 2, 3, or 5 until the desired quotient is an integer. In this example, you have: (30 ^ (20*x)) / ((20*x) ^ 30) As you increase x, the numerator will grow faster than the denominator and eventually the quotient will be an integer. In this case, x=2 fails, but x=3 works, x=5 works, x=6 works, etc. Tom
Excellent!
------ Quoting Tom Karzes <karzes@sonic.net>:
A can have prime factors that B lacks. Example: A=6, B=32.
Tom
I've been playing with commutators lately, and the following puzzle turned up ...
Suppose A and B are integers > 1.
When is A^B / B^A an integer? What values are possible?
Clearly A and B must have the same prime divisors, but the primes can occur to different powers. A = 45, B = 75 is a suggestive example.
Rich
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To be more precise, A^B/B^A is integer iff v_p(A)/A >= v_p(B)/B for each prime factor p of B. (Here v_p(x) denotes the p-adic valuation of x.) So scaling like Tom said will always work since v_p(B) grows much slower than B. Warut On Wed, Apr 28, 2010 at 7:11 AM, Tom Karzes <karzes@sonic.net> wrote:
In general, if you start with A and B, where A contains all of B's prime factors, then you can adjust B by scaling by A's factors. For example, if you start with A=30 and B = 20, you can keep scaling B by 2, 3, or 5 until the desired quotient is an integer. In this example, you have:
(30 ^ (20*x)) / ((20*x) ^ 30)
As you increase x, the numerator will grow faster than the denominator and eventually the quotient will be an integer. In this case, x=2 fails, but x=3 works, x=5 works, x=6 works, etc.
Tom
Excellent!
------ Quoting Tom Karzes <karzes@sonic.net>:
A can have prime factors that B lacks. Example: A=6, B=32.
Tom
I've been playing with commutators lately, and the following puzzle turned up ...
Suppose A and B are integers > 1.
When is A^B / B^A an integer? What values are possible?
Clearly A and B must have the same prime divisors, but the primes can occur to different powers. A = 45, B = 75 is a suggestive example.
Rich
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Warut Roonguthai