proove that the square of any positive integer of the form 3m+1 but nnot of the form 3q+2
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Hello Mate!!
Answer:-
Let 'a' be any positive integer.
On dividing it by 3 , let 'q' be the quotient and 'r' be the remainder.
Such that ,
a = 3q + r , where r = 0 ,1 , 2
When, r = 0
∴ a = 3q
When, r = 1
∴ a = 3q + 1
When, r = 2
∴ a = 3q + 2
When , a = 3q
On squaring both the sides,
a² = 9q²
a²= 3 × (3q)²
a²= 3
When, a = 3q + 1
On squaring both the sides ,
a² = (3q + 1) ²
a² = 9q² + 2 × (3q + 1) × 1²
a² = 9q² + 6q + 1
a² = 3 × (3q² + 2q ) + 1
a² = 3m + 1
where m = 3q² + 2q
When, a = 3q + 2
On squaring both the sides,
a² = ( 3q + 2 )²
a² = 3q² + 2 × 3q × 2 + 2²
a² = 9q² + 12q + 4
a² = ( 9q² + 12q + 3 ) + 1
a² = 3 × ( 3q² + 4q + 1 ) + 1
a² = 3m + 1
where m = ( 3q² + 4q + 1 )
Therefore, the square of any positive integer is either of the form 3m or 3m+1.
Hope it helps!!❤
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