MATHS……………………..?

Angel Saijal: MATHS……………………..?
show that square of any positive odd integer is of the form 8q+1 for some integer q.

plz i need explanation…………….

Answers and Views:

Answer by Michael jackson
oh..sorry,im very poor in math,u better ask this ur teacher

Answer by Geometry Warrior
[edit] ok, now i think i see what you meant there.
3^2 = 9
8(1) + 1 = 9

5^2 = 25
8(3) + 1 = 25

7^2 = 49
8(6) + 1 = 49

9^2 = 81
8(10) + 1 = 81

11^2 = 121
8(15) + 1 = 121, etc….

Has something to do with:
(odd number)^2 = 8(q) + 1
but the trick is figuring out the relationship between the odd number and q, and it’s just not coming to me in formula form.

Answer by Dibyajit d
the positive odd integer is in the form of (8q+1)
so its square will be 64q^2+1+16q=16q(4q+1)+1.
Here 16q(4q+1) is even and the addition of 1 makes it odd.
so the square is an odd integer

Answer by rishabh agarwal
For example you take 3.square of 3 is 9 and it is of form 8q+1,where q is 1.
Explanation——————————->
Every odd positive integer is of form (2n+1) ,where n ranges from o to infinity.Now when you square this term you get
(2n+1)^2=4n^2+1+4n.
This can be regrouped and can be written as
4n(n+1)+1——————————-(1),
where n ranges from 1 to infinity.Now if n is even or odd then n+1 will be odd or even and the product of odd and even is always even ,so we will be getting an even term.
now n(n+1) is always even we proved it.
Now every even number can be grouped as some 2*q
therefore,n(n+1)=2q.Now substitute this value in equation 1 ,then we will be getting as
(2n+1)^2=4*2q+1
therefore ,(2n+1)^2=8q+1.
Hence explained in detailed

Answer by $# ρ汉$$ΐ☼₪汉个€ ρυ₪נ汉ßΐ๏๏๏ #$
LET a be any positive integer and b = 8
therefore r= 0,1,2,3,4,5,6,7 (since 0<=r<b)
therfore a= bq+r
a= 8q,8q+1,…..8q+7
a (odd)=bq+1,8q+3..8q+7
a square= 64q square+1(m=8q square)
,,,,,=64q square +9+16*3q=8(8q square+24q)+8+1=8(8q square+24q+1)+1
understood

plz best awnser