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# 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

Combinatorics

8/10

##### Suggested Book
Challenges and Thrills in Pre College Mathematics

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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