Question

Show that the square of any positive integers is either of the form 3m or 3m + 1 for some ineteger m.

Answer 1

Answer:

Use Euclid division lemma to show that the square of any positive integer is either of the form 3m or 3m+1 for some integer m. Let a be any positive integer and b = 3. =) ... =) a = 3q or 3q + 1 or 3q + 2 for positive integer q.

Answer 2

Question :–

Show that the square of any positive integers is either of the form 3m or 3m + 1 for some ineteger m.

Solution :–

Let the required integer be,

$\boxed{ \sf \orange{a = bq + r}}$

ATQ,

$\boxed{ \sf \orange{a = 3q + r}} \: \boxed{ \sf \orange{r = 012}}$

Squaring both sides,

$\boxed{ \sf {a}^{2} = {(3q + r)}^{2} }$

$\implies {a}^{2} = {9q}^{2} + 6qr + {r}^{2}$

$\rule{200}3$

⓵ If r = 0

${a}^{2} = {9q}^{2} + 6q(0) + {(0)}^{2}$

$\implies {a}^{2} = {9q}^{2} = 3 {(3q}^{2} )$

$\implies {a}^{2} = 3m$

Where m = $\sf {3q}^{2}$ is another integers.

$\rule{200}3$

⓶ If r = 1

${a}^{2} = {9q}^{2} + 6q(1) + {(1)}^{2}$

$\implies {9q}^{2} + 6q + 1$

$\implies 3( {3q}^{2} + 2q) + 1$

$\implies 3m + 1$

Where m = $\sf {3q}^{2} + 2q$ is another integers.

$\rule{200}3$

⓷ If r = 2

${a}^{2} = {9q}^{2} + 6q(2) + {(2)}^{2}$

$\implies {9q}^{2} + 12q + 4$

$\implies {9q}^{2} + 12q + 3 + 1$

$\implies 3( {3q}^{2} + 4q + 1) + 1$

$\implies 3m + 1$

Where m = $\sf {3q}^{2} + 4q + 1$ is another integers.