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📅 May 12, 2026 👤 scraface

May 12, 2026

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📌 Each such trio corresponds to a 2-edge path (i.e., two adjacent edges sharing a vertex, not forming a triangle).

📌 For each internal node of degree at least 2 in the close pair graph, we can count how many such paths go through it.

📌 Let \( G \) be the graph with 5 vertices and 3 edges. Let \( d(v) \) be degree of \( v \).

📌 A path of length 2 is determined by a middle vertex and two neighbors.

📌 But we need the path to have *exactly* two edges in \( E \), and the three vertices involved fully described.

📌 So for each vertex \( v \), the number of paths of length 2 with center \( v \) is \( inom{\deg(v)}{2} \), and each such path \( u-v-w \) uses edges \( uv \) and \( vw \), both in \( E \). Then the trio \( \{u,v,w\} \) has exactly two close pairs (if no other edge), and since \( u \) and \( w \) are not necessarily connected, itâs exactly 2.

📌 But we must ensure that no additional edge exists, but since we are counting based on edge set, and the trio is defined by the three vertices, and the condition is on the number of close pairs among the three edges, if two are in \( E \) and one is not, then the trio has exactly two close pairs.

📌 So total number of such trios is the number of unordered triples \( \{u,v,w\} \) such that exactly two of the three pairs \( (u,v), (v,w), (u,w) \) are in \( E \).

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Wait — consecutive even integers: must be both even.
Wait — perhaps negative? But larger one positive likely.
Wait, 520 is sum of squares. Try solving new:
Wait — perhaps I misread. Try actual values:
No pair adds to 520? But problem says two consecutive even integers.
So no integer solution? But that cant be.
Wait — perhaps consecutive even integers and sum of squares 520 — lets...
n=16: 2(256)=512, 512 + 64 + 4 = 580
Wait — perhaps the sum is 180? But problem says 520.
Wait — perhaps 12 and 20? Not consecutive.

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