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