It isn’t the right answer to the question as posed. However it looks like (but isn’t necesarrily) the right answer to a very similar question, a third variant.

I would be very interested in seeing your working if you have time to write it up.

This is what I think you missed in the problem statement:

[spoiler]Although the co-efficients are constrained to be fractions there are no constraints on the test numbers.[/spoiler]

The actual answer relies on a neat bit of theory I intend to explain in a follow up post. It relates to: [spoiler]algebraic and trancendental numbers[/spoiler]

This is my analysis of the third variant, I will write it up properly in a follow up post.

[spoiler]If we constrain the test numbers to be integers or fractions then there is no solution. If we knew the degree of the polynomial, n, then we could do it in n+1 test numbers. Otherwise it takes an infinite number of guesses; there is a proof by contradiction as follows: Suppose the polynomial is p(x) and we can identify it with the test numbers t0, t1, t2, … , tm. Now put q(x) = (x-t0)(x-t1)…(x-tm). Clearly q(ti) will be zero for each of the test numbers. Put r(x) = q(x) + p(x). r(ti) = p(ti) for each of the test numbers so we cannot distinguish between r(x) and p(x). Therefore we have a proof by contradiction.[/spoiler]

I’ve just edited this because I’ve realised I’ve missed a case: [spoiler]what if the test numbers can only be algebraic numbers (E.g. 3, 47/236, sqrt(59) but not pi or e)? My proof by contradiction doesn’t work. Is the answer n+2? I need more time to think about it! Would love to see your working![/spoiler]

Thanks czarandy, that was fun!

It seems like you need n+2, if n is the degree of the polynomial.

[/spoiler] ]]>