Show that the square of an odd positive integer is of the form 3m+1, for any integer m.
Answers
Answer:
Any positive odd integer is of the form 2q + 1, where q is a whole number. q (q + 1) is either 0 or even. So, it is 2m, where m is a whole number. = 8 m + 1.
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 , a = 3q
On squaring both the sides
a^2 = 9q^2
a^3 = 3× (3q^2)
a^2 =3
where m = 3q^2
when ,a = 3q +1
On squaring both the sides
a^2 = (3q+1)
a^2 = 9q^2+2×3q×1+1^2
a^2 = 9q^2 +6q+1
a^2 = 3 ( 3q^2 + 2q)+1
a^2 = 3m+1
where m = 3q^2 +2q
when a =3q +2
on squaring on both the sides
a^2 = ( 3q+ 2)^2
a^2 = 3q^2 +2×3q×2+2^2
a^2 = 9 q^2 + 12 q +4
a^2 = ( 9q^2 + 12q+3 )+1
a^2 = 3 ( 3q^2 +4q+1)+1
a^2 = 3m+1
where m = 3q^2+4q+1.
hence , show that the square is the odd positive integer is the form 3m+1 .
hope this helps you
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