On division by gcd

Question

I am interested in the following problem:

Let $a_1,a_2,...,a_{p+1}$ be a sequence of distinct positive integers where $p$ is prime. Show that we can find two numbers from this sequence such that largest of them divided by their $GCD$ is $\geq p+1$.

Here's what I have tried so far... Suppose that this is not true, then for all $a_i, a_j$, we have $\max\{a_i,a_j\}\leq p\cdot \gcd(a_i, a_j)$. Now, let $\gcd$ be $d$. Then, we have $a_i=b_id, a_j=b_jd$.

I am not able to use the prime condition too! Any help will be highly appreciated.

Answer 1

This answer completes rtybase's answer, which finished off with

• if $\color{red}{\gcd(a_i,a_j)=d}>1$ and $p\mid d$ then ...

However, note $i$ and $j$ will be used as general indices here instead of the specific ones mentioned in the other answer. Also, assume the sequence values are sorted into increasing order, for simpler handling. First, here are $2$ lemmas I will use later.

Lemma #$1$:

If there's any factor $e \gt 1$ where $e \mid a_i$ for all $1 \le i \le p + 1$, then $e \mid d = \gcd(a_i,a_j)$ so, with $j \gt i$, the fraction $\frac{a_j}{d}$ would be the same as if $e$ was divided from both $a_i$ and $a_j$. As such, the maximum fraction value would be the same as if $e$ was first divided from all $a_i$. Note that related proofs about moving out a common factor among GCD elements are given at Prove that $(ma, mb) = |m|(a, b)\ $ [GCD Distributive Law].

Lemma #$2$:

For any sub-sequence of $t \ge 1$ integers, i.e., $a_{b_i}$ for increasing indices $b_i$ with $1 \le i \le t$, we have $L = \operatorname{lcm}(a_{b_1},a_{b_2},\ldots,a_{b_t}) \ge ta_{b_1}$. Consider the strictly decreasing sequence $\frac{L}{a_{b_i}}$ for $1 \le i \le t$. With each value being a positive integer, the largest (i.e., first) one must be $\ge t$. Thus, $\frac{L}{a_{b_1}} \ge t \;\to\; L \ge ta_{b_1}$. Note similar proofs for this are given at LCM of list of strictly increasing positive integers, and in the AoPS thread number theory.

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Use lemma #$1$ to divide by any common factor among all the $a_i$, so there's no factor $\gt 1$ in common among all the $a_i$. After doing this, if $p \mid d = \gcd(a_i,a_j)$ for some distinct $i$ and $j$, then there are

$$2 \le f \le p \tag{1}\label{eq1A}$$

sequence values which are a multiple of $p$ (note it's not all of them, i.e., $p + 1$, as discussed just above). Consider the indices of these values to be $g_i$, so we get for some integers $h_i$ that

$$a_{g_i} = h_i(p), \; 1 \le i \le f \tag{2}\label{eq2A}$$

Among the remaining

$$m = (p + 1) - f \tag{3}\label{eq3A}$$

sequence values which have no factor of $p$, if for any one of them, say $a_n$, for some $1 \le i \le f$ we have $\gcd(a_{g_i},a_n) = q \lt h_i$, then $\frac{a_{g_i}}{q} \ge 2p \gt p + 1$. If $a_n \gt a_{g_i}$, then $\frac{a_{n}}{q}$ would be even larger. In either case, we have a ratio $\ge p + 1$.

Otherwise, for all $n$ where $p \nmid a_n$, we have

$$\gcd(a_n,a_i)=h_i \;\;\forall\;\; 1 \le i \le f \tag{4}\label{eq4A}$$

Then, with $r = \operatorname{lcm}(h_1,h_2,\ldots,h_f)$, we get as shown in lemma #$2$ that

$$r \ge f(h_1) \tag{5}\label{eq5A}$$

Also, we must have $r$ dividing all of those $m$ sequence values with no factor of $p$. As such, the largest of these $m$ values, say $a_u$, has

$$a_u \ge mr \tag{6}\label{eq6A}$$

From \eqref{eq5A}, this means

$$a_u \ge mf(h_1) \;\;\to\;\; \frac{a_u}{h_1} \ge mf \tag{7}\label{eq7A}$$

Using \eqref{eq3A}, we get

$$mf = ((p + 1) - f)f = (p + 1)f - f^2 \tag{8}\label{eq8A}$$

If \eqref{eq8A} is $\ge p$, then because it's a lower bound in \eqref{eq7A}, it means we're finished. This is because, by using \eqref{eq2A}, with a lower bound of $p$ (i.e., for $f=p$), we have $\frac{a_u}{h_1} = p \;\to\; a_u = ph_1 = a_{g_1}$. However, since $u \neq g_1$, this contradicts the requirement the integers are all distinct, so this possibility is not valid. Thus, $\frac{a_u}{h_1}$ will have to be at least $p+1$ anyway. We can therefore use $i=g_1$ and $j=u$ because, from \eqref{eq2A} and \eqref{eq4A}, we have $h_1 = \gcd(a_u, a_{g_1})$.

For $2 \le f \le p - 1$, note that \eqref{eq8A} is always $\ge p + 1$. This can easily be shown since, if we let $w(x) = (p + 1)x - x^2$, then $w(x) = w(p + 1 - x)$, with $w(2) = w(p - 1) = 2p - 2$. Also, $w'(x) = p + 1 - 2x \ge 0 \implies x \le \frac{p + 1}{2}$. In other words, $w(x)$ increases up to $x = \frac{p + 1}{2}$ and then decreases symmetrically.

In summary, this handles all the remaining cases not covered by rtbyase's answer to prove the original requested problem always holds, i.e., there exist at least $2$ sequence values such that the ratio of the larger value to their $\gcd$ is always at least $p + 1$.

Answer 2

Partial answer

From the given conditions and using pigeonhole principle, there will $\exists a_i,a_j\in \{a_1,a_2,...,a_{p+1}\}$ s.t. $a_j\equiv a_i \pmod{p}$ (i.e. $p+1$ remainders all $< p$). Or simply $$a_j=k\cdot p +a_i \tag{1}$$ for a $\color{red}{k\geq 1}$ since all the $a_i$'s are distinct. In this configuration we assume $\max\{a_i,a_j\}=a_j$ of course, but this doesn't affect the final result. Now

• if $\gcd(a_i,a_j)=1 \Rightarrow \max\{a_i,a_j\}=a_j=\color{red}{k}\cdot p +a_i \leq p$ - contradiction.
• if $\color{red}{\gcd(a_i,a_j)=d}>1$ and $p\nmid d$ then, because $p$ is prime, $\color{red}{d\mid k}$ leading to $$\max\{a_i,a_j\}=a_j=k\cdot p +a_i \leq p\cdot d \iff \\ \color{red}{\frac{k}{d}}\cdot p + \color{red}{\frac{a_i}{d}}\leq p$$ which is another contradiction, given $\frac{k}{d}\geq 1$ and $\frac{a_i}{d}\geq 1$.
• if $\color{red}{\gcd(a_i,a_j)=d}>1$ and $p\mid d$ then ...

Answer 3

My attempt (mostly complimentary of rtybase's answer) :

First, consider arithmetic progressions with start element also the distance added. We see immediately, that if we factor out the common factor, we land at $1,\ldots ,p+1$; for which it's trivial (just take first and last elements).

If we continue, with arithmetic progressions where the start term is a factor of the distance, we get to a sequence with 1 in it and top term is necessarily $\geq p+1$ (for which it is again trivial).