Question

show that the cube of any positive integers of is of the form for M, 4 m + 1 or 4 m + 3 for some integer m​

Answer 1

$\huge{\rm{\underline{\underline{Question:-}}}}$

Show that the cube of any positive integer is of the form 4m, 4m + 1 or 4m + 3 for some integer m.

$\huge{\rm{\underline{\underline{Solution:-}}}}$

Let 'n' be any positive integer.

Applying Euclid's division lemma with divisor 4, we get -

n = 4q , 4q+1 , 4q+2 or 4q+3 , where q is some whole number.

Four cases arise:-

Case l - If n = 4q , then-

$\large{\rm{ {n}^{3} = {(4q)}^{3} = 64 {q}^{3 } = 4m ,}}$ where $\large{\rm{{16q}^{3} }}$ is an integer.

Case II - If n = 4q + 1, then-

$\large{\rm{ {n}^{3} = {(4q + 1)}^{3} }}$

$\large{\rm{ \: \: \: \: \: \:= {64q}^{3} + 3. {(4q)}^{2} .1 + 3.4q. {(1)}^{2} + {(1)}^{3}}}$

$\large{\rm{ \: \: \: \: \: \: = 64 {q}^{3} + 48 {q}^{2} + 12q + 1 }}$

$\large{\rm{ \: \: \: \: \: \: = 4(16 {q}^{3} + 12 {q}^{2} + 3q) + 1}}$

$\large{\rm{ \: \: \: \: \: \: = 4m + 1}}$ ,

where $\large{\rm{ m = 16 {q}^{3} + 12 {q}^{2} + 3q }}$ is an integer.

Case III - If n = 4q + 2 , then-

$\large{\rm{ {n}^{3} = {(4q + 2)}^{3} }}$

$\large{\rm{\: \: \: \: \: \: = {64q}^{3} + 3. {(4q)}^{2} .2 + 3.4q. {(2)}^{2} + {(2)}^{3}}}$

$\large{\rm{ \: \: \: \: \: \: = 64 {q}^{3} + 96 {q}^{2} + 48q + 8 }}$

$\large{\rm{ \: \: \: \: \: \: = 4(16 {q}^{3} + 24 {q}^{2} + 12q + 2) }}$

$\large{\rm{ \: \: \: \: \: \: = 4m }}$ ,

where $\large{\rm{ m = 16 {q}^{3} + 24 {q}^{2} + 12q + 2}}$ is an integer.

Case IV - If n = 4q + 3 , then-

$\large{\rm{ {n}^{3} = {(4q + 3)}^{3} }}$

$\large{\rm{\: \: \: \: \: \: = {64q}^{3} + 3. {(4q)}^{3} .2 + 3.4q. {(3)}^{2} + {(3)}^{3}}}$

$\large{\rm{ \: \: \: \: \: \: = 64 {q}^{3} + 144 {q}^{2} + 108q + 27 }}$

$\large{\rm{ \: \: \: \: \: \: = 4(16 {q}^{3} + 36 {q}^{2} + 27q + 6) + 3}}$

$\large{\rm{ \: \: \: \: \: \: = 4m + 3}}$ ,

where $\large{\rm{ m = 16 {q}^{3} + 36 {q}^{2} + 27q + 6}}$ is an integer.

Hence , the cube of any positive integer is of the form 4m , 4m + 1 or 4m + 3 for some integer m.

Answer 2

$\huge\mathfrak{\underline{\underline{Solution:-}}}$

Let the given positive integer be a which when divided by 4 gives q as the quotient and r as the remainder.

•°• by Euclid's Division Lemma, we have

a = 4q + r, where r is greater than or equal to 0 and less than 4.

Thus, the possible reminders are 0, 1, 2, and 3.

Hence, a could be

=> a = 4q

=> a = 4q + 1

=> a = 4q + 2

=> a = 4q + 3

Case 1 : When a = 4q

a³ = (4q³)

a³ = 64q³

a³ = 4(16q³)

a³ = 4m (where m = 16q³)

Case 2 : When a = 4q + 1

a³ = (4q + 1)³

a³ = 64q³ + 48q² + 12q + 1

a³ = 4(16q³ + 12q² + 3q) + 1

a³ = 4m + 1 (where m = 16q³ + 12q² + 3q)

Case 3 : When a = 4q + 2

a³ = (4q + 2)³

a³ = 64q³ + 96q² + 48q + 8

a³ = 4(16q³ + 24q² + 12q + 2)

a³ = 4m (where m = 16q³ + 24q² + 12q + 2)

Case 4 : When a = 4q + 3

a³ = (4q + 3)³

a³ = 64q³ + 144q² + 108q + 27

a³ = 4(16q³ + 36q² + 27q + 6) + 3

a³ = 4m + 3 (where m = 16q³ + 36q² + 27q + 6)

Thus, the cube of any positive integer is of the form 4m, 4m + 1 or 4m + 3 for some integer m.