2

From Putnam and Beyond. My question is, how does this proof work? I'd like to know more about the intuition behind its mechanics.

Prove that there is no polynomial $P(x) = a_n x^n + a_{n-1} x^{n-1} + \ldots + a_0 $, with integer coefficients and of degree at least 1 with the property that $P(0), P(1), P(2), \ldots ,$ are all prime numbers.

Solution:

Assume the contrary.
Let $P(0) = p$ with $p$ 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.

Winther
  • 24,478
odomojuli
  • 91
  • I'll edit my post for better clarity – odomojuli Sep 10 '16 at 22:31
  • I'm not sure what you're expecting. I think proofs by contradiction are counterintuitive by nature. And it's a pretty contrived problem to start with. – mathematician Sep 10 '16 at 22:41
  • What part of the proof is giving you trouble? Are you having trouble understanding it or are you asking how one might think to produce the proof? – wgrenard Sep 10 '16 at 22:44
  • 1
    @mathematician: why it is "pretty contrived" to ask the very natural question: can there be a polynomial function on the integers whose values are al prime? – Rob Arthan Sep 10 '16 at 23:11
  • The question is especially not "contrived" since there were many attempts to produce prime numbers with polynomials. For example, famous Euler's $P(n)=n^2 +n+41$ which gives prime numbers for naturals less than 40. – Ennar Sep 11 '16 at 00:04

3 Answers3

8

Notice that since $a_0=p$ we have that

$$P(kp)=a_nk^n\cdot p^n+a_{n-1}k^{n-1}\cdot p^{n-1}+\dots + p$$

In other words, for each $k \in \mathbb{N}$, $P(kp)$ is a (finite) sum of multiples of $p$ (notice the powers of $p$ on each term). Hence, $P(kp)$ is divisible by $p$ for all $k \in \mathbb{N}$.

However, by hypothesis $P(kp)$ is prime for all $k \in \mathbb{N}$ (because $kp \in \mathbb{N}$ for all $k \in \mathbb{N}$).

In other words, we have that, for each $k \in \mathbb{N}$, $P(kp)$ is a prime that is divisible by $p$ (which is itself a prime). This means $P(kp)=p$ for all $k \in \mathbb{N}$, so $P$ takes the same value (the value $p$) infinitely many times.

Do you understand why this last bit (taking a value infinitely many times, for a polynomial) is a contradiction?

Fimpellizzeri
  • 23,126
  • this last bit (taking a value infinitely many times, for a polynomial) is a contradiction It's not a contradiction unless you add (and use) the assumption that $P$ is of degree $\ge 1$. – dxiv Sep 10 '16 at 22:58
  • @dxiv: the question says the polynomial has degree at least 1. – Rob Arthan Sep 10 '16 at 23:13
  • 1
    @RobArthan Yes, but the answer as written does not make the point that that's an essential assumption. Given that the OP is asking for clarification of a given proof, I think it's important to carefully emphasize all the relevant points. – dxiv Sep 10 '16 at 23:16
  • 1
    I disagree. The answer above explains the key algebraic trick and asks the OP whether he or she needs more help to complete the proof. That's good pedagogy in my opinion. – Rob Arthan Sep 10 '16 at 23:33
  • @RobArthan We'll have to agree to disagree then. The part that's highlighted in my first comment is false as textually written, and may further confuse rather than help a poster who is admittedly fuzzy on some part of the proof. Of course, that's just my opinion, which everybody is free to ignore. – dxiv Sep 11 '16 at 00:04
2

There is an alternative proof provided by Hardy in his book "An Introduction To The Theory Of Numbers", Theorem 21, page 18 enter image description here

rtybase
  • 16,907
0

If the last part confuses you, you can proceed as follows:

Let $Q(x)=P(x)-p$.

Notice that $Q$ has the same degree as $P$ (say, $n$). Then $Q(x)=0$ infinitely many times, but this is a contradiction since $n$ is finite.

rowcol
  • 887