Question

Show that square of any positive integer can not be of the form 3m+2,where m is a natural number

Answer 1

According to Euclid division lemma
a=bq+r  where 0<= r < b
Therefore 0 = < r < 3
Therefore r= 0 , 1 , 2

If a=bq+r where r = 0
a=3q+0
squaring both sides
 a2= (3q+0)2
a2= 9q2
a2= 3×3× q2
3m where, m= 3q2

If a=bq+r where r = 1
a=3q+1
squaring both sides
a2= (3q+1)2
a2= 9q2+6q+1
a2= 3 (3q2+2q+1)
3m+1 where, m= 3q2 + 1

If a=bq+r where r =2
a=3q+2 
squaring both sides
a2= (3q+2)2
a2 = 9q​2 + 12q + 4
a2 = 3 (3q2 + 4q ) +4
a2= 3m+4

So, square of any poistive integer ccan't form 3m +2

Answer 2

Hey there !!

Let "a" be any positive integer , and 
b = 3q

here ,
r = 0 , 1 , 2

when r = 0
a = 3q
a² = (3q)²
    = 9q²
    = 3(3q²)
    = 3m    [ where m = 3q² ]                      ---> [1]

when r = 1
a   = 3q + 1
a² = ( 3q + 1 )²
    = 9q² + 6q + 1
    = 3 ( 3q² + 2q) + 1
    = 3m + 1    [ where m = 3q² + 2q ]         ---> [2]

where r = 2
a = 3q + 2
a² = (3q + 2 )²
    =9q² +12q + 4
    = 9q² +12q + 3 + 1
   = 3(3q² +4q + 1 ) + 1
   = 3 m +1   [ where m = 3q² +4q + 1  ]           ---> [3]


from 1 , 2 and 3 ,
its clear that the  square of any positive integer can not be of the form 3m+2,where m is a natural number