Question

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

Answer 1

we know that any positive odd integer is of the form 4q+1 or 4q+3.
let a be any positive integer,then:
case 1: a=4q+1
a^2= (4q+1)^2 = 16q^2+8q+1
= 8(2q^2+q)+1
= 8m+1 , m=(2q^2+q)

case 2: a=4q+3
a^2=(4q+3)^2 =16q^2+24q+9
= 8(2q^2+3q+1)+1
=8m+1 ,m=(2q^2+3q+1)

hence, square of any positive odd integer is of the form 8m+1

Answer 2

Step-by-step explanation:

Hey friends !!

[ Note :- I am taking q as some integer . ]

Let 'a' be the any positive integer.

Then, b = 8 .

Using Euclid's division lemma :-

0 ≤ r < b => 0 ≤ r < 8 .

•°• The possible values of r is 0, 1, 2, 3, 4, 5, 6, 7.

▶ Question said Square of odd positive integer , then r = 1, 3, 5, 7 .

→ Taking r = 1 .

a = bm + r .

= (8q + 1)² .

= 64m² + 16m + 1

= 8( 8m²+ 2m ) + 1 .

= 8q + 1 . [ Where q = 8m² + 2m ]

→ Taking r = 3 .

a = bq + r .

= (8q + 3)² .

= 64m² + 48m + 9 = 64m² + 48m + 8 + 1 .

= 8( 8m²+ 6m + 1 ) + 1 .

= 8q + 1 . [ Where q = 8m² + 6m + 1 ]

→ Taking r = 5 .

a = (8q + 5)² .

= 64m² + 80m + 25 = 64m² + 80m + 24 + 1 .

= 8( 8m²+ 10m + 3 ) + 1 .

= 8q + 1 . [ Where q = 8m² + 10m + 3 ]

→ Taking r = 7 .

a = ( 8q + 7 )² .

= 64m² + 112m + 49 = 64m² + 112m + 48 + 1 .

= 8( 8m²+ 14m + 6 ) + 1 .

= 8q + 1 . [ Where q = 8m² + 14m + 6 ] .

Hence, the square of any odd positive integer is of the form 8q + 1 .

✓✓ Proved ✓✓

THANKS

#BeBrainly.