Algebra Challenge – 2025/05/31

Claim: There is no $P(x)$ to satisfy $P(d)=\dfrac{2025}{d}$, where $d$ is a positive divisor of $2025$.

Lemma: If $P(x)$ is a polynomial with integer coefficients, and $a$ and $b$ are different integers. Then $P(a)-P(b)$ is a multiple of $a-b$.

Let $P(x)=\sum_{i=0}^{n}a_ix^i$. Then $P(a)-P(b)=\sum_{i=0}^{n}a_i(a^i-b^i)$. Since $a^i-b^i$ is a multiple of $a-b$, and $a_i$ are integers, therefore $P(a)-P(b)$ is a multiple of $a-b$.

Proof: Since $2025=3^4\cdot 5^2$, therefore $d=1, 3, 5, 9, 15, 25, 27, 45, 75, 81, 135, 225, 405, 675, 2025$.

Let $a=135$, $b=81$, we have $$P(a)=P(135)=\dfrac{2025}{135}=15$$ $$P(b)=P(81)=\dfrac{2025}{81}=25$$ $$P(a)-P(b)=15-25=10$$

However, $a-b=135-81=54$, and $10$ is not a multiple of $54$. Therefore, $P(x)$ does not exist.