Use Euclid's division lemma to show that the square of any positive integer is either of the form `3m`or `3m+1`for some integer m.[Hint: Let x be any positive integer then it is of the form `3q`, `3q+1`or `3q+2`Now square each of these and sho

Explanation

Let a be any positive integer .
Let `" "` b =3
By Euclid's division lemma
a = 3q + r
where , `0 le r lt 3 `
i.e., `" " ` r = 0, 1, 2
(i) When , r = 0 , then
a = 3q
`implies " " a^(2) = 9q^(2) = 3(3q^(2)) = 3m` , where m = `3q^(2)` is an integer .
(ii) When , r =1 , then
a = 3q +1
`implies " " a^(2) = (3q +1)^(2) = 9q^(2) + 6q + 1 = 3(3q^(2) + 2q) + 1 = 3m+1 `
where , `m = 3q^(2) + 2q` is an integer .
(iii) When r = 2 , then
`a - 3q + 2`
`implies " " a^(2) = (3q+2)^(2)`
=`9q^(2) + 12q + 4 = 3 (3q^(2) + 4q +1) + 1 = 3m+1 `
where , `m = 3q^(2) + 4q +1` is some integer .
Hence , the square of any positive integer is of the form 3m or 3m +1 for some integer m and no , it cannot be of the form 3m +2. `" "` Hence proved .