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