(In reply to

computer exploration by Charlie)

Fair disclosure: the
following was written** after seeing Charlieâ€™s results.**

Consider the function f =a+x^2.

It is easily shown that f
is
a multiple of 4a+1 both
for x=2a and x=2a+1.

Indeed :

a+(2a)^2=a*(4a+1)

a+(2a+1)^2=4a^2+5a+1= (a+1)*(4a+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.