First of all, I'm guessing you intended to write
a function from the integers to the integers.
Second, if
a constant, then from the given functional relation
it
follows that
, and this quadratic has two real non-integer solutions, contradiction.
Next, inputting
(a)
in the func. relation we get
;
(b) inputting
we get
, and;
(c) inputting
we get
Now you can find
as a function of
and substitute then in (a) to get
and etc. (Do you know/remember how to search for rational/integer roots of integral polynomials? This can come handy in this case)
Hmmm... the problem is that the only two possible values for
yield non-integer values for
...
.Check this.
I miscalculated: there's only one possible value for
which then gives one integer value for
and then one unique integer value for
...fine!
Tonio