00:06
We need to prove that the square of any integer is either in the form of square of any integer, in the form of 3k or 3k plus 1 and then we need to prove that the cube of any integer is 9k, 9k plus 1 or 9k plus 8.
00:44
So for the first case, the square case, we can write for case 1 if n is equal to 3k, then n squared that is equal to 9 k squared that is equal to 3 in bracket 3k square that is equal to 3m, then m is equal to 3k squared is an integer.
01:13
Therefore n square is the form of 3m where m belongs to z for case 2 if n is equal to 3k plus 1 then n square that is equal to 3k plus 1 whole square that is equal to 9 k square plus 6k plus 1 that is equal to 3 in bracket 3 k square plus 2k plus 1 that is equal to 3 l plus 1 where l is equal to 3k square plus 2 which is also an integer therefore n square is in the form of 3l plus 1 now in case 3k we can write n is equal to 3k plus 2 tk plus 2 whole square that is equal to 9 k square plus 12k plus 4 that is equal to 9 k square plus 12k plus 3 plus 4 that is equals to 3 in bracket 3k square plus 4k plus 1 plus 1 that is equal to 3p plus 1 where p is equal to 3k square plus 4k plus 1 and 3p plus 1 is also an integer therefore n square is in the form of 3p plus 1 so we can say that the square of any integer can be 3k or 3k plus 1.
02:53
Now we need to prove that the cube of any integer are in the form of 9k, 9k plus 1 or 9k plus 8.
03:05
So first of all, take a first case that is n is equals to suppose 3k.
03:12
So n cube that is equal to 3k, that is equal to 3k, that is equal...