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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