Define power(n) = log(n) / log N(n) , where N(n) is the conductor or square free core of n. Prove that power(n) = 1 if and only if n is square free and n is powerful implies power(n) >= 2

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- May 17th 2010, 11:58 PMSmithpower(n)
Define power(n) = log(n) / log N(n) , where N(n) is the conductor or square free core of n. Prove that power(n) = 1 if and only if n is square free and n is powerful implies power(n) >= 2

- May 18th 2010, 07:13 AMhollywood
I see the first part: n is squarefree and N(n)=n are essentially saying the same thing.

But the second part is not true: for n=548=137*2^2, N(n)=137, and power(n)=log(548)/log(137)=1.28....

- Hollywood - May 20th 2010, 06:42 PMthefrog
But Hollywood, 548 is NOT a powerful number; as you said 548 = 137*2^2, and a number is powerful if every prime in its prime factorisation appears to the power of 2 or a higher power, and 137 appears only to the power of 1 in the prime factorisation of 548.That invalidates your counterexample.

- May 21st 2010, 02:06 AMhollywood
You're right - I don't know what I was thinking.

If n is a perfect square, then log(N(n)) = log(1) = 0, so that power(n) is undefined. Otherwise, if n is powerful, it would seem that each prime in the prime factorization of n would add at least 3 times the value to the numerator as the denominator, so that power(n) is greater than or equal to 3.

- Hollywood - May 21st 2010, 05:40 PMhollywood
I was thinking that the condition was "n is not squarefree" instead of "n is powerful". Sloppy reading, I guess.....

So for some small non-squarefree (or "squareful") numbers (like for 12, power(n)=log(12)/log(3)=2.26...), power(n) is greater than two. You need to have a big prime factor to the first power to get power(n) down below 2. Actually, 20 = 5*2^2 is the first squareful number for which power(n) is less than 2, so I didn't need to go all the way up to 548.

- Hollywood