 Solution

An odd composite number $N$ with $\gcd (b,N)=1$ is a Fermat pseudoprime to base $b$ whenFor $N=35$, the Chinese remainder theorem for unit groups reduces this condition to congruences modulo $5$ and $7$. Modulo $5$, Fermat's little theorem givesso $b^{34}\equiv 1(\mathrm{mod}5)$ exactly when $b\equiv \pm 1(\mathrm{mod}5)$. Modulo $7$, the unit group has order $6$, soThe equation $b^4=1$ in the cyclic group $(\mathbb{Z}/7\mathbb{Z}{)}^{\times }$ has $\gcd (4,6)=2$ solutions, namely $b\equiv \pm 1(\mathrm{mod}7)$.

Combining the two independent sign choices by the Chinese remainder theorem givesand hence