partitioning 1 to p-1 into several a+b=c (mod p)

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partitioning 1 to p-1 into several a+b=c (mod p)

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#1 h Apr 21, 2024, 5:32 PM

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Given a prime number $p$ , a set is said to be $p$ -good if the set contains exactly three elements $a, b, c$ and $a + b \equiv c \pmod{p}$ .
Find all prime number $p$ such that $\{ 1, 2, \cdots, p-1 \}$ can be partitioned into several $p$ -good sets.

Proposed by capoouo

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#3 h Nov 21, 2024, 3:57 PM • 1 Y

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The answer is all $3 | p-1$ . Let $1, ..., p-1$ rewrite into $e, g, g^2, ..., g^{3k-1}$ . We can verify that $e + g^k + g^{2k} = 0$ as $g^{3k} = e$ , giving $g^{c} + g^{k+c} = g^{-2k-c} = g^{k-c}$ . As $c$ runs through $0, 1, ..., k-1$ we achieve $k$ $p-good$ sets. The contrary is obvious.

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#4 h Apr 8, 2025, 10:29 AM

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11 also satifsfies with sets $(1,3,9)$ , $(2,4,6)$ , $(5,7,8,10)$ can someone post a correct solution?

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#5 h Apr 8, 2025, 2:33 PM

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DTforever wrote:

11 also satifsfies with sets $(1,3,9)$ , $(2,4,6)$ , $(5,7,8,10)$ can someone post a correct solution?

the set has to contain exactly three elements

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