Euler's Phi Function

bluedevilgirl7 · #1 $Old$ November 4th, 2009, 11:53 PM

σ(n)Φ(n)≥n^2(1-1/p1^2)(1-1/p2^2)...(1-1/pr^2)

I can show that

σ(n)Φ(n)≥n^2(1-1/p1)(1-1/p2)...(1-1/pr)

but am unsure of how to do it with the pi^2.

tonio · #2 $Old$ November 5th, 2009, 05:52 AM

Quote:

Originally Posted by bluedevilgirl7 $View Post$

σ(n)Φ(n)≥n^2(1-1/p1^2)(1-1/p2^2)...(1-1/pr^2)

I can show that

σ(n)Φ(n)≥n^2(1-1/p1)(1-1/p2)...(1-1/pr)

but am unsure of how to do it with the pi^2.

I think it'd be a good idea if you defined what you use: what's that sigma function: the sum of all the divisors of the number n? And those p_i's are the prime divisors of n? If so the claim is false:

$\phi(6)=2\Longrightarrow 6\phi(6)=6\cdot 2=12\,,\,\,but\,\,6^2\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)=36\cdot\frac{3}{4}\cdot \frac{8}{9}=24$
By the way, without the squares we get equality with n = 6.

Tonio