Re: [數論] 找 n 使得 36|n^9+n^6+n^3+1
※ 引述《iddee ()》之銘言:
: 求所有小於 36 的正整數 n 使得 36 | (n^9 + n^6 + n^3 + 1)
12 3 9 6 3
n - 1 = (n - 1)(n + n + n + 1). Notice that
12
36∣n - 1 <=> g.c.d.(n,36) = 1.
9 6 3 6 3
Next, we see that n + n + n + 1 =(n + 1)(n + 1). If
6 3
g.c.d.(n,36) = 1, then n is odd, both n + 1 and n + 1 are even, and
6 3
so 4∣(n + 1)(n + 1).
3 6 6
Now, if 3∣n + 1, then 3∣n - 1, 3 cannot divide n + 1.
3
Thus 9∣n + 1, n ≡ 2, 5, 8 (mod 9). In summary,
n ≡ 2, 5, 8 (mod 9) and n ≡ 1, 3 (mod 4),
n = 29, 5, 17, 11, 23, 35.
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09/19 17:37, , 1F
09/19 17:37, 1F
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