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nick1816
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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: A

skmaqeel51
why do both have to be even?


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