Question

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.​

Answer 1

Step-by-step explanation:

Let a be any positive integer such that 0<r<=b

and EDL.

Let b=3 and a= 3q+r where 0<r<=3. i.e. r= 0,1,2

a= (3q). when r= 0

(a)= (3q)^2

a^2= 9q^2

a^2= 3(3q^2)

a^2= 3m where m= 3q^2

a= (3q+1). when r =1

(a)^2= (3q+1)^2

a^2= 9q^2+6q+1

a^2= 3(3q^2+2q) +1

a^2= 3m+1 where m= 3q^2+2q

a= 3q+2. where r= 2

a^2= 9q^2+12q+4

a^2= 9q^2+12q+3+1

a^2= 3(3q^2+4q+1)+1

a^2= 3m+1 where m= 3q^2+4q+1

Therefore, we can say that square of any positive integer is of the form 3m or 3m +1

Answer 2

Question: 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.​

Solution:

We'll prove this statement using Euclids Divison Lemma.

$\boxed{\begin{minipage}{10 cm}{\sf \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \underline{\underline{Euclids Division Lemma.}}}\\ \\ { \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \sf a = bq + r}\\ \\{\sf \ \ \ \ \ \ \ \ \ Where \ 'a' \ and \ 'b' \ are \ positive \ integers \ and \ 0 \leq \sf r < b.}\end{minipage}}$

$\sf \underline{\underline{Case \ I}}\\ \\\sf \Longrightarrow a=bq+r\\ \\\sf Let \ b=3 \ and \ r=0\\ \\\sf \Longrightarrow a=3q+0\\ \\\sf Squaring \ on \ both \ sides,\\ \\\sf \Longrightarrow (a)^2=(3q)^2+0\\ \\\sf \Longrightarrow a^2=9q^2\\ \\\sf \Longrightarrow a^2=3(3q^2)\\ \\\sf Let \ 'm' \ stand \ for \ 3q^2\\ \\\sf \Longrightarrow a^2=3m \ \ \ \longmapsto Eq(1)$

$\sf \underline{\underline{Case \ II}}\\ \\ \sf \Longrightarrow a=bq+r\\ \\\sf Let \ b=3 \ and \ r=1\\ \\\sf \Longrightarrow a=3q+1\\ \\\sf Squaring \ on \ both \ sides,\\ \\\sf \Longrightarrow a^2=(3q+1)^2\\ \\\sf \Longrightarrow a^2=9q^2+2(3q)(1)+1\\ \\\sf \Longrightarrow a^2=9q^2+6q+1\\ \\\sf \Longrightarrow a^2=3(3q^2+2q)+1\\ \\\sf {Let \ 'm' \ stand \ for \ 3q^2+2q}\\ \\\sf \Longrightarrow a^2=3m+1\ \ \ \longmapsto Eq(2)$

$\sf \underline{\underline{Case \ III}}\\ \\ \sf \Longrightarrow a=bq+r\\ \\\sf Let \ b=3 \ and \ r=2\\ \\\sf \Longrightarrow a=3q+2\\ \\\sf Squaring \ on \ both \ sides,\\ \\\sf \Longrightarrow a^2=(3q+2)^2\\ \\\sf \Longrightarrow a^2=9q^2+2(3q)(2)+2\\ \\\sf \Longrightarrow a^2=9q^2+12q+4\\ \\\sf \Longrightarrow a^2=9q^2+12q+3+1\\ \\\sf \Longrightarrow a^2=3(3q^2+4q+1)+1\\ \\\sf {Let \ 'm' \ stand \ for \ 3q^2+ 4q+1}\\ \\\sf \Longrightarrow a^2=3m+1\ \ \ \longmapsto Eq(3)$

From Equations 1, 2, and 3 we can say that the square of any positive integer is either the form of 3m, or 3m + 1, for some positive integer 'm'.