Hi skmaqeel51,Good question, because nick's line is a little compressed and it's easy to read it as "even x always gives even y." That's
not the rule. A polynomial like p(x) = x + 1 gives an odd y at x =
6. So let's pin down what's actually being claimed.
The real constraint is about the difference between the two y-values.With integer coefficients, write p(x) as a sum of terms c·x^k. Look at p(
6) - p(
4):
- Each term contributes c·(
6^k -
4^k).
- Every
6^k -
4^k is divisible by
6 -
4 =
2 (even minus even, and the pattern holds for any power).
- So the whole thing, p(
6) - p(
4), must be
even.
An even difference means p(
6) and p(
4) have the
same parity - either both even or both odd. That's the actual fact nick was leaning on.
Now apply it: the point (
6,
8) forces p(
6) =
8, which is
even. Same parity then forces p(
4) to also be
even. But the other point demands p(
4) =
9, which is
odd. Impossible - so
no such polynomial exists, and the answer is
A.
Quick checks to feel the rule:- p(x) = x + 1: p(
4) =
5, p(
6) =
7 - both odd (same parity).
- p(x) = x^2: p(
4) =
16, p(
6) =
36 - both even (same parity).
Every integer-coefficient polynomial does this: at two even inputs, the two outputs always match in parity. You can never get one even (
8) and one odd (
9) - which is exactly why the problem breaks.
Answer: Askmaqeel51
why do both have to be even?