# Set of primes $p_{1}\equiv 3 \bmod p_{2}$ such that $\phi(2^{\frac{{p_1}-3}{p_2}}-1)\equiv 0 \bmod p_1$ with $p_1,p_2\equiv 3\bmod 4$?

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

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