Question

show that cube of any positive integer is in the form of 4m, 4m+1,4m+2

Answer 1

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

Let the positive integer be a then, according to Euclid's division lemma

a = bq + r

Here,

b = 4

r = 0,1,2 and 3

Now,

a = 4q

a = 4q + 1

a = 4q + 2

a = 4q + 3

$\bold{\underline{In\ the\ 1st\ case\ we\ have\ a = 4q}}$

${a}^{3} = (4q) {}^{3} \\ \\ {a}^{3} = 64 {q}^{3} \\ \\ {a}^{3} = 4(16 {q}^{3} ) \\ \\ {a}^{3} = 4m$

Where m = 16${q}^{3}$

$\bold{\underline{In\ the\ 2nd\ case\ we\ have\ a = 4q + 1}}$

${a}^{3} =( 4q + 1 ) {}^{3} \\ \\ {a}^{3} = 64 {q}^{3} + 48 {q}^{2} + 12q + 1 \\ \\ {a}^{3} = 4(16 {q}^{3} + 12 {q}^{2} + 4q) + 1 \\ \\ {a}^{3} = 4m + 1$

where m = 16${q}^{3}$ + 12${q}^{2}$ + 4q

$\bold{\underline{In\ the\ 3rd\ case\ we\ have\ a = 4q + 2}}$

${a}^{2} = (4q+ 2) {}^{3} \\ \\ {a}^{3} = 64 {q}^{3} + 96 {q}^{2} + 48q+ 8 \\ \\ a {}^{3} = 4(16 {q}^{3} + 24 {q}^{2} + 12q + 2) \\ \\ {a}^{3} = 4(16 {q}^{3} + 24 {q}^{2} + 12q) + 2 \\ \\ {a}^{3} = 4m + 2$

where m = 16${q}^{3}$ + 24${q}^{2}$ + 12q

So, now the cube of positive integer a is in the form of 4m, 4m + 1 and 4m + 2

$\huge{\bold{\underline{Hence\ proved}}}$

Answer 2

Correct Question :-

Show that cube of any positive integer is in the form of 4m, 4m + 1, 4m + 3

Solution :-

We know that, a number is of the form of

a = bq + r, where a, b are positive integers and 0 ≤ r < b

Now, a number can be of the form of

a = 4q

a = 4q + 1

a = 4q + 2

or, a = 4q + 3

if, b is 4

So, for the first case :-

a = 4q

⇒ a³ = (4q)³

⇒ a³ = 64q³

⇒ a³ = 4(16q³)

⇒ a³ = 4m

where, m = 16q³

Second case :-

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)

Third Case :-

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)

Fourth Case :-

a = 4q + 3

⇒ a³ = (4q + 3)³

⇒ a³ = 64q³ + 576q² + 108q + 27

⇒ a³ = 64q³ + 576q² + 108q + 24 + 3

⇒ a³ = 4(16q³ + 144q² + 27q + 6) + 3

⇒ a³ = 4m + 3

where, m = (16q³ + 144q² + 27q + 6)

Hence Proved.