Question

show that the square of an odd positive integers is of the form 8m+1, for some whole number m.

Answer 1

Any odd positive integer is of the form 4q + 1 or 4q + 3 for some integer q. When n = 4q + 1, 
n²=(4q+1)²
   =16q² +8q+1
    = 8q(2q+1)+1
    =8m+1       where m = q(2q+1)
   
now n² is  of the form 8m+1
  

WHEN n = 8m+3
n²=(4q+3)²
   = 16q²+9+ 24q
  = 16q²+24q+8+1
  =8(2q²+3q+1)+1
  =8m+1                                where m = (2q²+3q+1)
  
n ² is of the form 8m+1 
HENCE
n² is of the form 8m +1 if n is an odd positive integer

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

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