Answer: A Polynomial Evaluated at Consecutive Integers

Key Idea / Intuition

If $p(n) = 0$ for some integer $n$, then $(x - n)$ divides $p(x)$ over the integers. This forces $p(k) = (k-n) \cdot q(k)$ for all integers $k$, where $q$ has integer coefficients. Among any two consecutive integers $k$ and $k+1$, one of $k - n$ and $k+1 - n$ is even — so at least one of $p(k), p(k+1)$ would be even. But we are told all 2025 values are odd — a contradiction.

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Formal Proof / Solution

Claim: $p$ has no integer roots.

Proof by contradiction. Suppose $n \in \mathbb{Z}$ is a root of $p$, so $p(n) = 0$.

Since $p$ has integer coefficients and $n$ is an integer root, the factor theorem over $\mathbb{Z}$ gives: $p(x) = (x - n)\, q(x)$ for some polynomial $q(x)$ with integer coefficients.

Now consider any integer $k$. We have: $p(k) = (k - n)\, q(k),$ where both $(k-n)$ and $q(k)$ are integers.

Key observation: Among any two consecutive integers $k$ and $k+1$, exactly one of the differences $(k - n)$ and $(k+1 - n)$ is even (since they differ by 1, they have opposite parity). Therefore, at least one of $p(k)$ and $p(k+1)$ is even.

Apply this to the consecutive pairs $(1,2),\, (2,3),\, \ldots,\, (2024, 2025)$. In each pair, at least one value is even.

But by hypothesis, $p(1), p(2), \ldots, p(2025)$ are all odd — so in particular, both elements of every consecutive pair are odd.

This is a contradiction.

Therefore, $p$ has no integer roots. $\blacksquare$

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Remark: The argument works for any set of consecutive integers of size $\geq 2$. Even two consecutive odd values of a polynomial (e.g., $p(1)$ and $p(2)$ both odd) is already enough to rule out integer roots. The number 2025 is irrelevant — even knowing $p(1)$ and $p(2)$ are both odd suffices!