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Answers & Comments
Jeevaking
Let the positive integer be 'a'. to prove, a²=3m or 3m+1 let a be divided by 3 =>b=3 =>r=0,1,2 a=bq+r (by remainder theorem) a=3q+r a=3q,3q+1,3q+2 a²=(3q)²=9q²=3(3q²)=3m [m=3q²] a²=(3q+1)²=9q²+6q+1=3(3q²+2q)+1=3m+1 a²=(3q+2)²=9q²+12q+4=9q²+12q+3+1=3(3q²+4q+1)+1=3m+1
therefore, square of any positive integer is of the form 3m or 3m+1.
Answers & Comments
to prove,
a²=3m or 3m+1
let a be divided by 3
=>b=3
=>r=0,1,2
a=bq+r (by remainder theorem)
a=3q+r
a=3q,3q+1,3q+2
a²=(3q)²=9q²=3(3q²)=3m [m=3q²]
a²=(3q+1)²=9q²+6q+1=3(3q²+2q)+1=3m+1
a²=(3q+2)²=9q²+12q+4=9q²+12q+3+1=3(3q²+4q+1)+1=3m+1
therefore, square of any positive integer is of the form 3m or 3m+1.
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