Question

Number Theory

Consider a polynomial P(x) = (x + d1)(x + d2) · . . . · (x + d9), where d1, d2, . . . , d9 are nine
distinct integers. Prove that there exists an integer N such that for all integers x ≥ N the
number P(x) is divisible by a prime number greater than 20.

Answer 1

Sorry let me write it better.

Answer 2

Consider a polynomial P(x) = (x + d1)(x + d2) · . . . · (x + d9), where d1, d2, . . . , d9 are nine distinct integers. Prove that there exists an integer N such that for all integers x ≥ N the number P(x) is divisible by a prime number greater than 20.

Answer 3

They give me this hints.

Note that the statement of the problem is invariant under translations of x; hence
without loss of generality we may suppose that the numbers d1, d2, . . . , d9 are positive.

The key observation is that there are only eight primes below 20, while P(x) involves more
than eight factors.

Answer 4

Considering that the polynomial must be divisible by a prime, it is assumed that one of the roots must be prime, or a factor of a prime. Since that there is nine roots in the polynomial, and that all integers are positive and different(which means the roots are all different), then there must at most be nine different prime number roots, and since there are only eight distinct prime integers below 20, it is shown at least one of them must be a prime above 20.

Answer 5

Here is the proof that the teacher gave us.

We shall prove that N = d8 satisfies the desired property, where d = max{d1, d2, . . . , d9}. Suppose for the sake of contradiction that there is some integer x ≥ N such that P(x) is
composed of primes below 20 only. Then for every index i ∈ {1, 2, . . . , 9} the number x + di can be expressed as product of powers of the first 8 primes. Since x + di > x ≥ d8 there is some prime power fi > d that divides x + di. Invoking the pigeonhole principle we see that there are two distinct indices i and j such that fi and fj are powers of the same prime number. For reasons of symmetry, we may suppose that fi ≤ fj. Now both of the numbers x + di and x + dj are divisible by fi and hence so is their difference di − dj . But as 0 < |di − dj | ≤ max(di, dj) ≤ d < fi

Answer 6

Wow. Quite vigorous, I see. lol Using the Reductio ad absurdum