Question

Show that square of an odd +ve integer is of the form 8m+1for some integer m

Answer 1

HEY MATE ✌✌✌

Answer:

Step-by-step explanation:

By Euclids division lemma, given positive integers a and b, there exist unique integers q and r satisfying a = bq + r, 0 ≤ r < b.

a = 4q + r where 0_< r<4

So r may be 0, 1, 2, 3

putting 1 in place of r

a=4q+1

a^2= (4q+1)^2

a^2=16q^2+8q+1

a^2=8q(q+1) + 1

a^2= 8m+1 where 'm' is equal to q(q+1)

hence square of a positive odd integer is of form 8m +1 for some integer 'm'

^_^^_^ HOPE IT HELPS ^_^^_^

Answer 2

Any odd no can be written in the form 2n+1

Its square is (2n+1)^2= 4n^2+4n+1
=4n(n+1)+1 ——>(1)
n(n+1) is always an even number as either n or n+1 will be an even no

So n(n+1) can be written as 2m

So (1) becomes 4*2m+1

=8m+1

Hence answered