Satis- $\phi$ !

We can apply difference of squares a bunch of times to get all the roots. $0=(1+ \phi)^4 -\phi^4 = ((1+ \phi)^2 -\phi^2) ((1+ \phi)^2 +\phi^2)$ $=((1+ \phi) -\phi) ((1+ \phi) +\phi) ( (1+ \phi) -i\phi) ((1+ \phi) +i\phi)$

Which will give 3 distinct solutions

Okay.

*Approach 1: De-Moivre's theorem *

We first manipulate the equation to our desired form as follows:

$\begin{aligned} \phi^4 &= (1+ \phi)^4 \\ \Rightarrow 1 &= \left( 1 + \dfrac{1}{\phi}\right)^4 \\ \Rightarrow \sqrt[4]{1} &= \left( 1 + \dfrac{1}{\phi}\right) \end{aligned}$

Now, We find the Complex Roots of the left hand side.

$\Rightarrow \left( e^{2k\pi i} \right)^\frac{1}{4} = \left( 1 + \dfrac{1}{\phi} \right)$ where $k = 0,1,2,3$

Thus, by substituting values for $k$ we get:

$\left( 1 + \dfrac{1}{\phi} \right) = 1 , i , -1 , -i$

Now, from the following values, we can find $\phi$ with only three values except $1$ as the Right Hand Side (RHS)

Thus only three such $\phi$ s exist:

$\begin{aligned} RHS = i &\rightarrow& &\phi =\dfrac{-i-1}{2} \\ RHS = -i &\rightarrow& &\phi =\dfrac{i-1}{2} \\ RHS = -1 &\rightarrow& &\phi = \dfrac{-1}{2} \end{aligned}$

Approach 2 : Factorization

Since $- \frac{1}{2}$ is a root, and that $f(x) = ( x+1)^4-x^4$ is a cubic polynomial, so we can factorize $f(x) = (2x+1)( 2x^2+2x+1)$ , and solve the quadratic.

Credit: Calvin Sir

Student taught by maths student won't get this wrong.

[(p + 1)^2 - p^2][(p + 1)^2 + p^2] = 2 (2 p + 1)(p^2 + p + 1/ 2) = 0

=> p = - 0.5 or p = - 0.5 + j 0.5 or p = - 0.5 - j 0.5

There are 3 complex numbers on complex field.