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let $p_1$ and $p_2$ be positive primes such that $p_1,p_2 \equiv 3\bmod 4$ and $\phi$ is the Euler totiont function , I want to find the Set of primes $p_{1}\equiv 3 \bmod p_{2}$ such that $\phi(2^{\frac{{p_1}-3}{p_2}}-1)\equiv 0 \bmod p_1$ ? I can't find such pairs satisfying the titled claim , I have used the fact that : $\phi(2^{\frac{{p_1}-3}{p_2}}-1)\equiv 0 \bmod \frac{{p_1}-3}{p_2}$ but this dosn't give any thing to determine all pairs $(p_1,p_2)$ for which the titled claim w'd be satisfied .

Edit I have Edited the question to avoid any complication and working with ordinary prime which are $3$ modulo $4$ in the same time gives the definition of Gaussian primes

Note:The motivation of this question is to know more about behavior of Euler totiont function with Gaussian primes

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    $\begingroup$ If $\frac{|z_1|-1}{|z_2|}$ is an integer, then $|z_{1}|\equiv 1 \bmod|z_{2}|$. $\endgroup$ – Wojowu Apr 11 at 16:33
  • $\begingroup$ primes of which form? $\endgroup$ – Fedor Petrov Apr 11 at 21:41
  • $\begingroup$ @FedorPetrov primes of the form $z= b i$ (Gaussian primes) , I have montioned that $\endgroup$ – zeraoulia rafik Apr 11 at 21:56
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    $\begingroup$ Why not just ordinary primes congruent to 3 mod 4, but multiplied by $i$? $\endgroup$ – Fedor Petrov Apr 11 at 22:02
  • $\begingroup$ Multiplied by i beacuse I meant Gaussian prime and it works also as you claimed $\endgroup$ – zeraoulia rafik Apr 11 at 22:04

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