Understand the problem

Suppose 91 distinct positive integers greater than 1 are given such that there are at least 456 pairs among
them which are relatively prime. Show that one can find four integers a, b, c, d among them such that
gcd(a, b) = gcd(b, c) = gcd(c, d) = gcd(d, a) = 1.

Source of the problem
Regional Math Olympiad, 2019 Problem 6

Topic
Combinatorics

Difficulty Level

8/10

Suggested Book
Challenges and Thrills in Pre College Mathematics

Start with hints

Do you really need a hint? Try it first!

The clue: Consider the numbers as points in a graph, and some edges between the points based on their gcd.

Let us consider a graph G with 91 vertices (91 distinct numbers) and connect each pair of these vertices. which corresponds to co-prime pairs implies at least 456 edges i.e. e \( \geq\) 456. Here e = number of edges. Now getting four number a,b, c, d such that gcd (a, b) = gcd (b, c) = gcd(c,d) = gcd (d,a) = 1 implies existence at a cycle of length 4.

 

We will follow the contradiction method.

Let us assume there is no cycle of length ‘4’. Our aim is to find a bound on e, which contradicts the 456 bound on e.

Let vertex set of G be V = {\(v_1, v_2,…,v_{91} \)}, let degree of \( v_i = d_i\).

Now, observe that the number of vertex pairs {\( v_k, v_l\)}, adjacent to \(v_i\) is \({d_i}\choose{2}\).

Now observe that a vertex pair {\( v_k, v_l\)} cannot be common to two vertices say \(v_i\) and \(v_j\) both, because if so, \( v_i, v_k, v_j, v_l, v_i\) will form a cycle.

So, the sum of all such vertex pairs corresponding to a vertex over all the vertex set \( \leq\) the total number of vertex pairs.

\( \sum_{i=1}^{91} ({d_i}\choose{2}) \leq {{91}\choose{2}} \rightarrow \)

\( \sum_{i=1}^{91} ({d_i}^2 – d_i)\leq 91.90 \rightarrow \)

\( \sum_{i=1}^{91} {d_i}^2 \leq 2e + {{91}\choose{2}}\) -> (1)

as \( \sum_{i=1}^{91} d_i = 2e \).

We will try to find a lower bound on \( \sum_{i=1}^{91} {d_i}^2 \) w.r.t e.

By RMS – AM inequality,

\( \sum_{i=1}^{91} {d_i}^2 \geq \frac{( \sum_{i=1}^{91} {d_i})^2 }{91} = \frac{4e^2}{91} \) -> (2)

Using (1) and (2), we get \( \frac{2e^2}{91} – e \leq 91.45 \)

From here, by solving we get \( e \leq 91×5 =455\). This gives the contradiction.

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