Question

show that the square of any positive odd integer is of the form 6m+1 or 6m+3

Answer 1

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Answer 2

Step-by-step explanation:

Let a be the square of any positive odd integer and b = 6 and r=0,1,2,3,4,5:

So, by Euclid's division lemma. we get that:

a=bq+r

a=(6q+0)²

a= 36q²

a= 6(6q)²

 = 6m (where 6q² is m)

a=bq+r

a=(6q+1)²

a= 36q²36q^2+12q+1

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

 = 6m+1 (where 6q^2+2q is m)

a=bq+r

a=(6q+2)²

a=36q^2+24q+4

a=6(6q^2+4q+2)+2

 = 6m+2 (where 6q^2+4q+2 is m)

a=bq+r

a=(6q+3)²

a=36q^2+36q+6+3

 =6(6q^2+6q+1)+3

 = 6m+3 (where 6q^2+6q+1 is m)

a=bq+r

a=(6q+4)²

a=36q^2+48q+16

 =36q^2+48q+12+4

 = 6(6q^2+8q+2)+4

 = 6m+4 (where 6q^2+8q+2 is m)

a=bq+r

a=(6q+5)²

a=36q^2+60q+24+1

 =6(6q^2+10q+4)+1

 = 6m+5 (where 6q^2+10q+4 is m)

Now, we need to remove 6m and 6m+2 and 6m+4 only because these are clearly divisible by 2 , hence, they are positive even integers.

So, it is proved that the square of any positive odd integer is of the form 6m+1 or 6m+3 or 6m+5.

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