Find N such that for all x $\ge$ N, $P(x)$ is Divisible by Primes Bigger than 10

Question

Given an arbitrary distinct natural numbers, $d_1, d_2, d_3, d_4,$ and $ d_5$.

Let $P(x) = (x + d_1)(x + d_2)(x + d_3)(x + d_4)(x + d_5)$.

Prove that there is a number $N$ (in terms of $d_1, d_2, ...$) such that for all $q \ge N$, $P(q)$ is divisible by prime bigger than 10.

I tried letting $P(x) = 2^a 3^b 5^c 7^d$ and consider $d_1, d_2, ...$ by their distance, but got stuck.

Answer 1

Solution via small cases and wishful thinking (where if things don't work out, we tweak them as needed.)

First, some observations

1. We need $d_i$ distinct, since otherwise $d_i = 0$ (or a constant) leads to no such $N$.
   a. This suggests we should look at $d_i - d_j$, or something similar.
2. There are 5 linear terms for us to avoid 4 primes.
   a. My current guess is that this is important. Maybe we need more linear terms than primes.
   b. It could still be worthwhile to consider if we could use 2 linear terms to avoid 2+ primes, in case the 5 linear terms is a red herring.

Wishful thinking 1: Show that for distinct $d_1$, $P(x) = (x+d_1)$ is eventually not of the form $2^a$.

Non-Proof: Whoa, this statement is clearly not possible. We will always have that form for $ x = 2^a - d_1$.

This suggests that Observation 2a comes into play, though it's not immediately clear as yet how it is involved.

Wishful thinking 2: Show that for distinct $d_1, d_2$, $P(x) = (x+d_1) ( x+d_2)$ is eventually not of the form $2^a$.

Proof: If $ (x+d_1)(x+d_2) = 2^a$, then $x+d_1 = 2^{a_1}, x+d_2 = 2^{a_2}$ and $|d_1 - d_2 | = |2^{a_1} - 2^{a_2}|$ gives us a limit on how large $a_i$ can be, and hence $2^a$. Thus, there is a large enough $N$ such that for all $q \geq N$, $P(q)$ is divisible by a prime $\geq 3$. $_\square$

Great, we seem to be making some progress.
It might be helpful to find a condition for such $N$. A sufficient condition is that $ \min (x+d_i) > |d_1 - d_2|$. EG For $ d_1 = 4, d_2 = 8$, we need to use $ x +4 > 4$.

Wishful thinking 3: Show that for distinct $d_1, d_2$, $P(x) = (x+d_1) ( x+d_2)$ is eventually not of the form $2^a3^b$.

Non-Proof: If $(x+d_1)(x+d_2) = 2^a$, then $x+d_1 = 2^{a_1}3^{b_1}, x+d_2 = 2^{a_2}3^{b_2}$ and $|d_1 - d_2 | = 2^{\min (a_1, a_2)}3^{\min(b_1, b_2)} \times C$ where $ C$ is a positive integer. This gives us a limit on $ \min(a_1, a_2) \leq \log_2 | d_1 - d_2|, \min(b_1, b_2) \leq \log_3 | d_1 - d_2|$.

It's not immediately clear to me how we can continue here. If the minimum applied to the same index (which need not always be the case), then we would have $x+d_i = 2^{a_i} 3^{b_i} < D^2$.
We only have bounds on the minimum values, but not the maximum values. (I don't want to rely on Pillai's conjecture about the distance between prime powers to get more information out of $C$.)
Recall that observation 2a suggests we should look at 3 linear terms and 2 primes, so let's do that.

Wishful thinking 4: Show that for distinct $d_1, d_2, d_3$, $P(x) = (x+d_1) ( x+d_2)(x+d_3)$ is eventually not of the form $2^a3^b$.

Proof: Let $D = \max | d_i - d_j|$. We repeat the above set up, and conclude that $ D \geq 2^{\min (a_i, a_j) } 3^{ \min (b_i, b_j) } C$. This means that we have a bound on all except 1 of $a_i$, and a bound on all except 1 of $b_i$. Since we have 3 terms, we thus have a bound on (at least) one of the $x+d_i$. In particular, when $\min(x+d_i) > D^2$, the condition will be satisfied. $_\square$

It's now clear how the observations come into play.

Wishful thinking 5: Show that for distinct $d_1, d_2, d_3, d_4, d_5$, $P(x) = (x+d_1) ( x+d_2)(x+d_3)(x+d_4)(x+d_5)$ is eventually not of the form $2^a3^b5^c7^d$.

Proof: Left to the reader.
Show further that when $N > \max|d_i - d_j|^4 - \min(d_k)$, the condition will be satisfied.

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I would love to see a proof or refutation of Wishful Thinking 3.

Answer 2

In fact, we can generalize this claim somewhat.

Theorem: suppose we have primes $p_1, \ldots, p_n$ and distinct integers $d_1, \ldots, d_n, d_{n + 1}$. Then for all sufficiently large $x$, at least one of $x + d_1, \ldots, x + d_{n + 1}$ has a prime factor other than $p_1, \ldots, p_n$.

Proof: let $L = \max\limits_{1 \leq i, j \leq n + 1} |d_i - d_j|$. For each $j$ between $1$ and $n$, choose $k_j$ such that $p_i^{k_j} > L$. Define $M = 1 + (\prod\limits_{j = 1}^n p_j^{k_j}) - \min\limits_{1 \leq i \leq n + 1} d_i$.

Now suppose that $x \geq M$. Suppose on the contrary that for all $1 \leq i \leq n + 1$, we can write $x + d_i = \prod\limits_{j = 1}^n p_j^{m_{i, j}}$. For each $i$, since $\prod\limits_{j = 1}^n p_j^{m_{i, j}} = x + d_i > \prod\limits_{j = 1}^n p_j^{k_j}$, we can choose some $\ell_i$ with $1 \leq \ell_i \leq n$ such that $m_{i, \ell_i} > k_{\ell_i}$.

By the pidgeonhole principle, there must exist $i_1 \neq i_2$ such that $\ell_{i_1} = \ell_{i_2}$; let $\ell = \ell_{i_1} = \ell_{i_2}$. Then we see that $p_\ell^{k_\ell}$ is a factor both of $x + d_{i_1}$ and $x + d_{i_2}$, so $p_\ell^{k_\ell}$ is a factor of $d_{i_1} - d_{i_2}$, hence a factor of $|d_{i_1} - d_{i_2}|$. However, we know that $0 < |d_{i_1} - d_{i_2}| \leq L < p_\ell^{k_\ell}$; contradiction. $\square$

To apply this theorem to our case, we work with $n = 4$ and primes $2, 3, 5, 7$.