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Answer 1

Question:

To show square of positive integer cannot be in the form of 6m+5

Answer:

                    let 'a' be positive integer.Thus,according to Euclid's division lemma

  a = 6q+r , where b=6 and 0$\leq r >6$

  $\implies r= 0,1,2,3,4,5.$

first case , let r = 1

$a=6q+1$

square both sides ,

a² =(6q++1)²

⇒a² = 36q²+12q+1

       =6(6q²+2q)+1

       =6m+1 , where 'm' = 6q²+2q

second case , r = 2

⇒ a = 6q+2

square on both sides,

⇒ a²  = (6q+2)²

⇒ a²  = 36q²+24q+4

        = 6(6q²+4q)+4

        = 6m+4 , where m = 6q² +4q

Third case, r = 3

⇒ a = 6q+3

square on both sides

⇒ a² = (6q+3)²

⇒ a² = 36q²+36q+9

        = 36q² +36q+6+3

       = 6(6q²+6q+1)+3

       = 6m+3 , where m = 6q²+6q+1

similarly you can do for the rest of the values of 'r'