Verified answer

Corrected version. Sorry for my earlier gaffe.

x = p1^n1 * ... pr^nr.

Then

φ(x) = p1^(n1-1) * (p1-1)...pr^(nr-1)*(pr-1) = 2p.

If all the pi are odd, this requires

p1^(n1-1)* (p1-1) = 2p (*)

because if r > 1, the product of the pi-1 would be

divisible by 4.

In this case, n1 = 1 or 2 and if n1 = 2 then

p1*(p1-1) = 2p. (*)

This implies p1 = 2 or p1 = p.

The first case is impossible, and if p1 = p

then p-1 = 0(mod 3).

But 2p ≡ 2(mod 3).

So 3 divides the left side of (*), but not the right side,

and this is impossible.

We are left with x = p1.

So p1-1 = 2p, p1 = 2p+1.

Since p ≡ 1(mod 3), we must have

p1 ≡ 0(mod 3), i.e., p1 = 3.

But then p-1 = 2, not 2p.

Finally, we may have p1 = 2 and the rest of the pi odd.

Then r = 2, because if r > 2 then the product of the

pi would be divisible by 4.

So x = 2^n1* p2^n2.

and φ(x) = 2^(n1-1)*p2^(n2-1)(p2-1) = 2p.

This yields n1 = 1 because p2-1 is even.

Thus this case reduces to the previous case.

Zayin W--This is the sort of problem you study in

elementary number theory.

properly the question being asked isn't an equation.An equation is like concerning something to a distinctive element.Like 3x=9(linear),ax^2 + bx +c = 0 (quadratic),ax^3 + bx + c = 0 (cubic) , x^2 + y^2 = 9 (u get a circle if u graph it.) (4p/p^2 - 2p -3) * (p-3 / p+one million) =[4p/(p-3)(p+one million)] *(p-3 / p+one million) =4p/(p+one million)^2

Kind off sorry this is not an answer actually a question!!

But What level Math is this??

what?

i'm lost. lmao