Without calculating them determine whether $36^2+1$ and $154^2+1$ are prime and find the prime factors if not prime I know that $36^2 + 1$ is prime, $154^2 + 1$ is not, both are equal to $1 \bmod 4$. The prime divisors of $154^2 + 1$ should also be of the form $1 \bmod 4$. Tried showing this by Wilson's theorem however I don't feel like I'm getting any where. 

The square roots of $-1 \pmod 5$ are $2,3 \pmod 5$ but $154 \equiv 4 \pmod 5.$
The square roots of $-1 \pmod {13}$ are $5,8 \pmod {13}$ but $154 \equiv 11 \pmod {13}.$
The square roots of $-1 \pmod {17}$ are $4,13 \pmod {17}$ but $154 \equiv 1 \pmod {17}.$
The square roots of $-1 \pmod {29}$ are $12,17 \pmod {29}$ but $154 \equiv 9 \pmod {29}.$
The square roots of $-1 \pmod {37}$ are $6,31 \pmod {37}.$  $$154 \equiv 6 \pmod {37}.$$
