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

Answer:

I HOPE THIS MAY HELP YOU

Answer 2

$\huge{\mathfrak{\green{\red{\bf\:S}\pink{o}\purple{l}\green{u}\orange{t}\blue{i}\blue{o}\red{n:-}}}}$

Let x be positive integers

• b = 3

By Euclid's division algorithm,

a = bq + r (where 0 ≤ or < b)

x = 3q + r (where 0 ≤ or < 3)

possible value of r = 0,1,2

$\underbrace{\bf\:Case\:1}$

when r = 0

then,

x = 3q

Squaring both side,

$\rm\longrightarrow\:{(x)}^{2}={(3q)}^{2}$

$\rm\longrightarrow\:{x}^{2}={9q}^{2}$

$\rm\longrightarrow\:{x}^{2}=3(3q^2)$

$\rm\longrightarrow\:{x}^{2}=3m\:\:\:\:\:\:[\therefore\:m=3q^2]$

$\underbrace{\bf\:Case\:2}$

When r = 1

then,

x = 3q + 1

squaring both side,

$\rm\longrightarrow\:{(x)}^{2}={(3q+1)}^{2}$

$\rm\longrightarrow\:{x}^{2}={(3q)}^{2}+2\times\:3q\times\:1+(1)^2$

$\rm\longrightarrow\:{x}^{2}={9q}^{2}+6q+1$

$\rm\longrightarrow\:{x}^{2}=3({3q}^{2}+2q)+1$

$\rm\longrightarrow\:{x}^{2}=3m+1\:\:\:\:\:\:[\therefore\:m=3{q}^{2}+2q]$

$\underbrace{\bf\:Case\:3}$

when r = 2

then,

x = 3q + 2

Squaring both side,

$\rm\longrightarrow\:{(x)}^{2}={(3q+2)}^{2}$

$\rm\longrightarrow\:{x}^{2}={(3q)}^{2}+2\times\:3q\times\:2+(2)^2$

$\rm\longrightarrow\:{x}^{2}=9{q}^{2}+12q+4$

$\rm\longrightarrow\:{x}^{2}=9{q}^{2}+12q+3+1$

$\rm\longrightarrow\:{x}^{2}=3(3{q}^{2}+4q+1)+1$

$\rm\longrightarrow\:{x}^{2}=3m+1\:\:\:\:\:\:\:[\therefore\:m=3{q}^{2}+4q+1]$

Hence,the square of any two positive integers is either of the form 3m or 3m + 1 for some integers m.