English Deutsch Français Italiano Español Português 繁體中文 Bahasa Indonesia Tiếng Việt ภาษาไทย
All categories

If n^2≠n mod 3, then n ≡ 2 mod 3.

2007-10-23 14:37:34 · 2 answers · asked by bandit8711 1 in Science & Mathematics Mathematics

2 answers

For any integer n there are three mutually exclusive possibilities:
n is congruent to 0 (mod 3)
n is congruent to 1 (mod 3)
n is congruent to 2 (mod 3)

If n is congruent to 0 (mod 3), square both sides of the congruence.
n^2 is congruent to 0^2 = 0 (mod 3)
Thus n^2 is congruent to n (mod 3)

If n is congruent to 1 (mod 3), square both sides again.
n^2 is congruent to 1^2 = 1 (mod 3)
Thus n^2 is congruent to n (mod 3) again.

Now if n is congruent to 2 (mod 3) and we square both sides,
n^2 is congruent to 2^2 = 4 which is congruent to 1 (mod 3).
But if n is congruent to 2 (mod 3) and n^2 is congruent to 1 (mod 3) then n^2 is not congruent to n (mod 3).

So n^2 is not congruent to n (mod 3) iff n is congruent to 2 (mod 3)

2007-10-23 14:57:18 · answer #1 · answered by wild_turkey_willie 5 · 2 0

There are only three possibilities: n≡0, n≡1, or n≡2 mod 3. But 0²≡0 and 1²≡1 mod 3, so if n≡0 or n≡1 mod 3, then n²≡n mod 3. Since we are told that n²≢n mod 3, that can't happen, so it must be the sole remaining possibility, namely that n≡2 mod 3.

2007-10-23 14:58:15 · answer #2 · answered by Pascal 7 · 0 0

fedest.com, questions and answers