# 3n+1_conjecture 8, n^n – 1 ≡ 0 mod 3, nのn乗から1を引いた整数が、3で割りきれるときのnを求めてみましょう。

Now, Let’s preparatory exercise for the solution of 3n + 1 conjecture today too. Today is an interesting expression … multiples of 3 and -1(in negative multiples) are fused to catch an even number.

-1(負の倍数)が、3で割り切れる作用に、偶数を捉えてきました。そんな感じです。

Q: Can you find the natural number(n) that satisfies below?
nn – 1 ≡ 0 mod 3

First, it transforms as follows about “nn – 1 ≡ 0 mod 3″.
nn ≡ 1 mod 3
nn ≡ 1n mod 3 ∨ nn ≡ (-1)n mod 3 (n is an even number) — (1)

From the relation between congruence expression and polynomial with integer coefficients,
(1) can be transformed into the following.
n ≡ 1 mod 3 ∨ n ≡ -1 mod 3 (n is an even number) — (2)

Then, (2) can be transformed into the following.
It defines k, m in non-negative integer.
n = 3k + 1 (k≧0) ∨ n = 6m + 2 (m≧0) — (3)

By the way, “n = 3k + 1” is “n = 6k + c (c is either 1 or 4)”.
Therefore, (3) can be transformed into the following.
n = 6m + c (m≧0, c is either 1, 2 or 4)

Therefore, the solution to be obtained is
“n = 6m + c (m is non-negative integer, c is either 1, 2 or 4)”.