Question

Prove that {}^{2} - 2" align="absmiddle" class="latex-formula">
will never be divisible by 3.​

Answer 1

Answer:

Any integer x can be represent in form of 3k, 3k+1, or 3k + 2 (k is integer).

Case 1: x = 3k => x^2 - 2 = 9k^2 - 2, which will not be divisible by 3

(notice that 2 is not divisibble by 3, but 9k^2)

Case 2: x = 3k + 1 => x^2 - 2 = 9k^2 + 6k + 1 - 2 = 9k^2 + 6k - 1, which will not be divisible by 3

(notice that 1 is not divisibble by 3, but 9k^2 + 6k)

Case 3: x = 3k + 2 => x^2 - 2 = 9k^2 + 12k + 4 - 2 = 9k^2 + 12k - 2, which will not be divisible by 3

(notice that 2 is not divisibble by 3, but 9k^2 + 12k)

=> x^2 - 2 will never be divisible by 3

Hope this helps!

:)