Show every prime factor of $4t^2 + 1$ is equivalent to 1 modulo 4
My working so far:
I want to use the first Nebensatz, so given q is a prime factor I want to show $(-1/q)=(-1)^{(q-1)/2}=1$ as this implies $q \equiv1 (mod 4)$
It would be sufficient to prove $x^2=-1 (mod 4t^2 +1)$ has solutions as we could then use Chinese remainder to imply the above, however I'm not sure how to prove this has solutions.
Occasionally, it is easier to (see how to) prove a stronger assertion than the required assertion.
If we try to find a solution of the equation $x^2 + 1 = 4t^2 + 1$ rather than a solution of the congruence $x^2 \equiv -1 \pmod{4t^2+1}$, the solution $x = 2t$ immediately jumps out. Hence a fortiori $(2t)^2 \equiv -1 \pmod{q}$ for every prime factor $q$ of $4t^2+1$, which shows $q\equiv 1 \pmod{4}$.