In other words, $z^n \equiv \bar{z}\pmod{n}$ or $z^{n+1} \equiv z\bar{z}\pmod{n}$. That splits into two conditions: $$\begin{cases} \Im z^{n+1}\equiv 0\pmod{n}, \\ (c^2+1)^{n}\equiv c^2+1\pmod{n}. \end{cases} $$

Numbers $n$ satisfying the first condition Grau et al. (2014) called Gaussian Fermat pseudoprimes to base $z$, while numbers satisfying the second condition (under the additional constraint $\gcd(c^2+1,n)=1$) are Fermat pseudoprimes to base $c^2+1$. There are many composite examples satisfying either of the two conditions, it is hard to find those that satisfy both.

For example, Grau et al. showed that there no composite $n=4k+3$ below $10^{18}$ that are both Gaussian Fermat pseudoprimes to base $1+2I$ and Fermat pseudoprimes to base $2$.

Alternatively, it can be seen that $z^n$ can be expressed in terms on Lucas sequences: $$z^n = \frac{1}{2}V_n(P,Q) + I U_n(P,Q)$$ for $(P,Q)=(2c,c^2+1)$, and hence $n$ satisfying the two conditions is a Frobenius pseudoprime.

In other words, $z^n \equiv \bar{z}\pmod{n}$ or $z^{n+1} \equiv z\bar{z}\pmod{n}$. That splits into two conditions: $$\begin{cases} \Im z^{n+1}\equiv 0\pmod{n}, \\ (c^2+1)^{n}\equiv c^2+1\pmod{n}. \end{cases} $$

Numbers $n$ satisfying the first condition Grau et al. (2014) called Gaussian Fermat pseudoprimes to base $z$, while numbers satisfying the second condition (under the additional constraint $\gcd(c^2+1,n)=1$) are Fermat pseudoprimes to base $c^2+1$. There are many composite examples satisfying either of the two conditions, but it is hard to find those that satisfy both.

For example, Grau et al. showed that there no composite $n=4k+3$ below $10^{18}$ that are both Gaussian Fermat pseudoprimes to base $1+2I$ and Fermat pseudoprimes to base $2$.

Alternatively, it can be seen that $z^n$ can be expressed in terms on Lucas sequences: $$z^n = \frac{1}{2}V_n(P,Q) + I U_n(P,Q)$$ for $(P,Q)=(2c,c^2+1)$, and hence $n$ satisfying the two conditions is a Frobenius pseudoprime.

In other words, $z^n \equiv \bar{z}\pmod{n}$ or $z^{n+1} \equiv z\bar{z}\pmod{n}$. That splits into two conditions: $$\begin{cases} \Im z^{n+1}\equiv 0\pmod{n}, \\ (c^2+1)^{n}\equiv c^2+1\pmod{n}. \end{cases} $$

Numbers $n$ satisfying the first condition Grau et al. (2014) called Gaussian Fermat pseudoprimes to base $z$, while numbers satisfying the second condition (under the additional constraint $\gcd(c^2+1,n)=1$) are Fermat pseudoprimes to base $c^2+1$. There are many composite examples satisfying either of the two conditions, it is hard to find those that satisfy both.

For example, Grau et al. showed that there no composite $n=4k+3$ below $10^{18}$ that are both Gaussian Fermat pseudoprimes to base $1+2I$ and Fermat pseudoprimes to base $2$.

In other words, $z^n \equiv \bar{z}\pmod{n}$ or $z^{n+1} \equiv z\bar{z}\pmod{n}$. That splits into two conditions: $$\begin{cases} \Im z^{n+1}\equiv 0\pmod{n}, \\ (c^2+1)^{n}\equiv c^2+1\pmod{n}. \end{cases} $$

Numbers $n$ satisfying the first condition Grau et al. (2014) called Gaussian Fermat pseudoprimes to base $z$, while numbers satisfying the second condition (under the additional constraint $\gcd(c^2+1,n)=1$) are Fermat pseudoprimes to base $c^2+1$. There are many composite examples satisfying either of the two conditions, it is hard to find those that satisfy both.

For example, Grau et al. showed that there no composite $n=4k+3$ below $10^{18}$ that are both Gaussian Fermat pseudoprimes to base $1+2I$ and Fermat pseudoprimes to base $2$.

Alternatively, it can be seen that $z^n$ can be expressed in terms on Lucas sequences: $$z^n = \frac{1}{2}V_n(P,Q) + I U_n(P,Q)$$ for $(P,Q)=(2c,c^2+1)$, and hence $n$ satisfying the two conditions is a Frobenius pseudoprime.