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Problem- Prove that there is no polynomial

$$P(x) = a_nx^n + a_{n−1}x^{n−1} +\dots+ a_0$$

with integer coefficients and of degree at least 1 with the property that $P (0), P (1), P (2)$, ... are all prime numbers.

Solution- Assume the contrary and let $P (0) = p, p$ prime.

Then $a_0 = p$ and $P (kp)$ is divisible by $p$ for all $k \geq 1.$

Because we assumed that all these numbers are prime, it follows that $P (kp) = p$ for $k \geq 1$.

Therefore, $P (x)$ takes the same value infinitely many times, a contradiction. Hence the conclusion.

I have some trouble understanding the last part of the solution. How does contradiction take place?

saulspatz
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Gnómi
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2 Answers2

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If a polynomial $P$ has degree $n$, then the equation $P(x)=k$ has, at most, $N$ solutions, for any fixed number $k$. But it was shown in the proof that the equation $P(x)=p$ has infinitely many solutions. Hence, there is a contradiction.

José Carlos Santos
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Consider the polynomial $P(x)-p$. The degree of this polynomial is same as that of $P(x)$.

The Fundamental Theorem of Algebra (FTA) guarantees:

An $n$-degree polynomial has atmost $n$ real roots.

But, as shown in your proof, $P(x)-p$ has an infinite number of roots of the form $kp$ (because, $P(kp)-p=p-p=0$), contradicting the FTA.

Hence, the result follows

Soham
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