(In reply to computer exploration
Fair disclosure: the
following was written after seeing Charlie’s results.
Consider the function f =a+x^2.
It is easily shown that f
a multiple of 4a+1 both
for x=2a and x=2a+1.
Since 4a+1 is a common
divisor and a dominant factor (in the only valid general factorization)
it is a g.c.d.
The 4a+1 is a basic solution
and any member of the sequence 2a+401*k will do.
Maybe that a formal proof requires showing the impossibility of getting a g.c.d. greater than 4a+1,
so I did not post it as a solution, it is rather an observation.