STEP I 1992 question 7 solution

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Solution 1

and

As n, g(0) and g(1) are integers, g(n) must be an integer too.

and

and

and

So

(I is the integer form of a)

We've now got integer forms of an^2, bn and c so f(n) can be expressed as an integer:

All of which are integers. A more complete f(n):

And by the by, in case someone wants a justification of why is an integer:

The left hand side is an integer. This implies that the right hand side is an integer. As I can take out a factor of 2 out of the right hand side, it implies that the right hand side is an even integer. This then implies that the left hand side is an even integer. And so the left hand side is divisible by 2.

Solution by DeathAwaitsU.

Solution 2

if g(1) and g(0) for any given value of n, g(n) is an integer.

Therefore because f(-1),f(0) and f(1) are integers

Now for the bit that harder

Take the first term Aalpha^2

next term -2A\alpha n

Next two easier terms Ba + C

Next term is Bn

So!

As \alpha and all the functions involving alpha are integers therefore f(\alpha + n) must be an integer.

Solution by insparato.