Question

show that square of any odd integer is form of 6 m + 1 or 6 m + 3 for some integer m​

Answer 1

Answer:

Hii friend,

Let n be a given positive odd integer.

On dividing n by 6 , let m be the Quotient and r be the remainder.

Then, by Euclid division lemma, we have

Dividend = Divisor × Quotient + Remainder

n = 6m + r. where r = 0 , 1 ,2 , 3 ,4 ,5

n= 6m + 0 = 6m. [ r =0]

n = 6m + 1 = 6m+1 [ r = 1 ]

n = 6m +2 = 6m +2 [ r = 2 ]

n = 6m +3 = 6m+3 [ r = 3 ]

n = 6m +4 = 6m+4 [ r = 4 ]

n = 6m+5 = 6m+5 [ r = 5 ]

N = 6m , (6m +2) , (6m+4) is even value of n.

Therefore,

when n is odd , it is in the form of (6m+1) , (6m+3) , (6m+5) for some integer m.

HOPE IT WILL HELP YOU..... :-)

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Answer 2

HIII....

HERE IS YOUR ANSWERE....

Let the positive integer be 'a'

by EDL

a=bq+r

=> b=3

=> 0=r<3

=> r=0,1,2,

=> a=6q,6q+1,6q+2

here a is odd no.

=> a= 6q+1

case 1

a=6q+1

S.B.S

a^2 =(6q+1)^2

=>a^2=36q^2+1+12q

= 6(6q^2+2q)+1

=> m= 6q^2+2q

=> a=6m+1

hence proved..

plzzz.. mark me as brainliest

hope this helps you