Linear Independence of Angles Corresponding to Primes $p \equiv 1 \pmod 4$ Any prime $p \equiv 1 \pmod 4$ can be written uniquely as the sum of two squares $p = a^2 + b^2$. In the complex plane, this corresponds to the 8 Gaussian integers with norm $p$. We can write these points as
$$
\sqrt{p}e^{2\pi i (\pm\Phi + t/4)}
$$
where we define $\Phi$ as the argument of the point in the first octant ($\Phi = \arg(a+bi)$ where $a^2 + b^2 = p$ and $a \geq b$) and $t \in \{0,1,2,3\}$ corresponds to multiplication by a unit.
I want to show that any finite set of the type $\left\{ 1, p_1, \ldots , p_N \right\}$ where $p_j \equiv 1 \mod 4$, $j = 1, \cdots, N$ is linearly independent over $\mathbb{Q}$. In other words, we want to show
$$
\sum_{j=1}^{N} \alpha_j \Phi_j + \alpha_0 = 0
\iff
\alpha_0 = \alpha_1 = \cdots =\alpha_N = 0
$$
where the $\Phi_j$ is defined for each $p_j$ as described above.
I encountered this problem while self-studying some number theory and have been stuck for a bit. I'd appreciate some hints! The approach I tried is to go by contradiction and try to show that something will contradict the definition of a prime number, but I think I've gotten lost!
 A: There was a previous answer here from @TheoBendit (I suppose he deleted it) but that was enough when combined with a hint from Ramin Takloo-Bighash to help me solve this.
Since we can multiply through by a common denominator, we can assume all the $\alpha_j$ are integers.
Let $a_j + ib_j = \sqrt{p_j}e^{2\pi i \Phi_j}$ (These coefficients come from $p_j = a_j^2 + b_j^2$) Suppose the angles are linearly dependent, then we can calculate
\begin{align*}
\prod_{j=1}^{n} (a_j + ib_j)^{2\alpha_j}
&=
\prod_{j=1}^{n} \left(\sqrt{p_j} e^{2\pi i \Phi_j} \right)^{2\alpha_j} \\
&=
\prod_{j=1}^{n} p_j^{\alpha_j} e^{2\pi i (-2\alpha_0)} \\
&=
\prod_{j=1}^{n} p_j^{\alpha_j} \\
&=
\prod_{j=1}^{n} (a_j + ib_j)^{\alpha_j} (a_j - ib_j)^{\alpha_j}
\end{align*}
Cancelling terms from both sides, we have
\begin{align*}
\prod_{j=1}^{n} (a_j + ib_j)^{\alpha_j}
=
\prod_{j=1}^{n} (a_j - ib_j)^{\alpha_j}
\end{align*}
This is a contradiction since $\mathbb{Z}[i]$ is a UFD.
Credit to Theo for motivating me to consider a product of this type, and credit to Ramin for the hint of raising everything to the $2\alpha_j$.
