Question

use euclids division lemma to show that the square of any positive integer is in the form of 3n or 3n+1 for any integer n

Answer 1

Let a be any positive integer and b=3

By euclid's division lemma,

a=bq+r, 0=, <r <b
a=3q+r

a=3q
a=3q+1
a=3q+2
now square each of each

when r=0,

a^2=(3q)^2
a^2=9q^2
3n=3 (3q^2)

n=3q^2


when r=1


a^2=(3q+1)^2
a^2=(3q)^2+2×3q×1+1^2
a^2=9q^2+6q+1
3n+1=3q (3q+2)+1
n =(3q+2)


when r=2

a^2=(3q+2)^2
a^2=(3q)^2+2×3q×2+2^2
a^2=9q^2+12q+4
3n+1=3(3q^2+4q +1)+1
n=3q^2+4q +




hope this helps you. ......





a^2=(3q

Answer 2

Heyaa folk
___________________________________________

you asked good question

Let's solve it



Let x be any positive integer and b that is quotient = 3

for remainder there is possible value are -. 0,1,2 { since 0 <_ r<b}
so we can write in the form of

x = 3q or 3q+1 or 3q+2

we apply division lemma
that is a = bq+ r

★ CASE 1 - If x = 3q + 0

x^2 = ( 3q)^2

x^2 = 9q^2

x2 = 3(3q^2)

= 3n where n = 3q^2

★ CASE 2 - If x = (3q+1)

x^2 = ( 3q+1) ^2

=. 9q^2 + 6q + 1 { since (a + b)^2 = a^2 + 2ab +b^2}

= 3 q( 3q +2) +1

= 3 n+1 where n = q ( 3q+2)

★CASE 3. -. If x = ( 3q+2)

x^2 =. (3q+2)^2

=. 9q^2 + 12q. + 4

= 3 ( 3q^2+ 4q +1) +1 { since we splitted 4 into form 3 + 1}

= 3 n+1 where n = ( 3q^2 + 4q + 1)

◆ HENCE THE SQUARE OF ANY POSITIVE INTEGER IS OF THE FORM OF 3n or 3n+1

REGARDS
#KUSHIK12




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