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• ... phi(n^2)=n*phi(n), for any n. so boils down to showing that phi(n)=n-1 iff n is prime. which is easily known. Andy
Message 1 of 4 , May 3 12:01 PM
> phi( (p^2) ) = [phi(p)]^2 + phi(p) iff "p" is prime,
> and phi(...) is the totient function.
>
> Or is this equation already known? Bill.

phi(n^2)=n*phi(n), for any n. so boils down to showing
that phi(n)=n-1 iff n is prime. which is easily known.

Andy

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• phi(p^2) = p*( p-1)......(1) phi(p)=p-1..........(2) Using 1 and 2 in phi( (p^2) ) = [phi(p)]^2 + phi(p) we get p*(p-1) = (p-1)^2 + p-1 I think the result is
Message 2 of 4 , May 3 5:46 PM
phi(p^2) = p*( p-1)......(1)
phi(p)=p-1..........(2)

Using 1 and 2
in
phi( (p^2) ) = [phi(p)]^2 + phi(p)
we get
p*(p-1) = (p-1)^2 + p-1

I think the result is obvious.........for the if
case.

Now for the only if case

Let N = p(1)^a(1) * p(2)^a92)*........*p(n)^a(n)

phi(N) =
p(1)^{a(1)-1}*p(2)^{a(2)-1}*......p(n)^{a(n)-1}*{p(1)-1}*{p(2)-1}.....{p(n)-1}

N^2= p(1)^2a(1) * p(2)^2a(2)*........*p(n)^2a(n)

phi(N^2) =
p(1)^{2a(1)-1}*p(2)^{2a(2)-1}*......p(n)^{2a(n)-1}*{p(1)-1}*{p(2)-1}.....{p(n)-1}

Now if the equation holds then

phi( (p^2) ) = [phi(p)]^2 + phi(p)

phi(N^2) = phi(N)^2 + phi(N)
==>
[p(1)^{2a(1)-1}*p(2)^{2a(2)-1}*......p(n)^{2a(n)-1}*{p(1)-1}*{p(2)-1}.....{p(n)-1}]
=
[p(1)^{a(1)-1}*p(2)^{a(2)-1}*......p(n)^{a(n)-1}*{p(1)-1}*{p(2)-1}.....{p(n)-1}]^2
+
[p(1)^{a(1)-1}*p(2)^{a(2)-1}*......p(n)^{a(n)-1}*{p(1)-1}*{p(2)-1}.....{p(n)-1}]

Cancelling off the common terms from both the sides
(implies none of p(i) = 1 or 0 )

The equation yields

p(1)^a(1) * p(2)^a92)*........*p(n)^a(n)=
p(1)^{a(1)-1}*p(2)^{a(2)-1}*......p(n)^{a(n)-1}*{p(1)-1}*{p(2)-1}.....{p(n)-1}
+ 1

N = phi(N) + 1

Or phi(N) = N-1
which is true iff N = prime

I think the proof is over
and the statement is very correct.

Shiv Nandan Singh

Shiv Nandan Singh

--- William Bouris <melangebillyb@...> wrote:
> Hello Group...
>
> Can someone prove or disprove?
>
> phi( (p^2) ) = [phi(p)]^2 + phi(p) iff "p" is prime,
> and phi(...) is the totient function.
>
> Or is this equation already known? Bill.
>
>
> -

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