Is there a field $K$ and a quadratic polynomial $f\in K[T]$ such that $f^2+1$ is a square in $K[T]$?

Question

Is there a field $K$ (ideally a number field not containing $i$) such that there is a quadratic polynomial $f = T^2 + aT + b\in K[T]$ satisfying the property that $f^2+1$ is also a square in $K[T]$?

For various complicated reasons my investigations has led me to consider this problem...

Answer 1

Not if $K$ has characteristic $\ne 2$.

If $f^2+1=g^2$, then $1=(f+g)(f-g)$ has degree $0$, but at least one of $f\pm g$ has degree $\ge 2$.

Answer 2

Of course:

$$f(t)=t^2\in\Bbb F_2[t]\implies f^2+1=t^4+1=(t^2+1)^2\in\Bbb F_2[t]$$

As a rule: where squares kick in in "tricky" ways, characteristic $\;2\;$ is what will likely render results

Answer 3

You can't do it unless char $K=2$.

Simply compare coefficients of $f^2+1=(T^2+cT+d)^2$. You get $2a=2c$, $a^2+2b=c^2+2d$, and you can divide by 2 to conclude $a=c$ and $b=d$, but then the identity can not hold.

If char $K=2$, it is always true: $f^2+1=(f+1)^2$.