Question

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

Answer 1

Let a be positive integer and b=2. Then by E.D.A

a=4q+r where 0<r<4

a^3= 64q^3+r^3+48q^2r+12qr^2

Case 1 =

a^3=64q^3

a^3=4(16q^3)

a^3=4m where m=16q^3

Case 2 = when r=1

a^3= 64q^3+48q^2+12q+1

a^3=4(16q^3+12q^2+3q)+1

a^3=4m+1 where m=16q^3+12q^2+3q

Case 3= When r=2

a^3=64q^3+96q^2+48q+8

a^3=4(16q^3+19q^2+12q+2)a^3=4m where m=16q^3+19q^2+12q+2

Case4when r=3

a^3=64q^3+144q^2+108q+27

a^3=4(16q^3+36q^2+27q+6)+3

a^3=4m+3 where m= 16q^3+36q^2+27q+3

it will surely help u

Answer 2

Let us take any two positive integer (say 'a' and 'b') where b = 4 and q is any natural number.

According to Euclid's Division Lemma

a = bq + r

Here, b = 4

So, a = 4q + r

0 ≤ r < b

0 ≤ r < 4

So, r = 0,1,2,3

Case (1):

When r = 0

a = 4q

Cubing both Sides,

${a}^{3} = 4q$

$\implies {a}^{3} = 64{q}^{3}$

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

$\implies {a}^{3} = 4m (where\;m = 16{q}^{3} )$

Case (2):

When r = 1

a = 4q + 1

Cubing both Sides,

${a}^{3} = {(4q +1)}^{3}$

$\implies {a}^{3} = 64{q}^{3} + 1 + 32{q}^{2} + 12q$

$\implies {a}^{3} = 4(16{q}^{3} + 8{q}^{2} + 3q) + 1$

$\implies {a}^{3} = 4m + 1 (where\;m = 16{q}^{3} + 8{q}^{2} + 3q)$

Case (3):

When r = 2

a = 4q + 2

Cubing both Sides,

${a}^{3} = {(4q +2)}^{3}$

$\implies {a}^{3} = 64{q}^{3} + 8 + 96{q}^{2} + 48q$

$\implies {a}^{3} = 4(16{q}^{3} + 2 + 24{q}^{2} + 12q )$

$\implies {a}^{3} = 4m (where\; m = 16{q}^{3} + 2 + 24{q}^{2} + 12q )$

Case 4:

When r = 3

a = 4q + 3

Cubing both Sides,

${a}^{3} = {(4q +3)}^{3}$

$\implies {a}^{3} = 64{q}^{3} + 27 + 144{q}^{2} + 108q$

$\implies {a}^{3} = 64{q}^{3} + 24 + 3 + 144{q}^{2} + 108q$

$\implies {a}^{3} = 4(16{q}^{3} + 6 + 36{q}^{2} + 27q ) + 3$

$\implies {a}^{3} = 4m + 3 (where\;m = 16{q}^{3} + 6 + 36{q}^{2} + 27q )$

Hence, the cube of any positive integer is in the form 4m, 4m+1 and 4m+ 3