Show that the square of an odd positive integer is of the form 8q + 1, for some integer q.
Answers
SOLUTION :
Since any odd positive integer n is of the form 4m + 1 and 4m + 3
Case : 1
If n = m + 1, then
n² = (4m + 1)²
[On squaring both sides]
n² = (4m)² + 8m + 1²
[(a+b)² = a² + b² + 2ab]
n² = 16m² + 8m + 1
n² = 8m (2m + 1) + 1
n² = 8q +1 , where q = m (2m + 1)
Case : 2
If n = 4m + 3, then
n²= (4m + 3)²
[On squaring both sides]
n² = (4m)² + 24m + 3²
[(a+b)² = a² + b² + 2ab]
n² = 16m² + 24m + 9
n² = 8 (2m² + 3m + 1) + 1
n² = 8q + 1 , where q = 2m² + 3m + 1
Hence, n² is of the form 8q + 1, for some integer q.
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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 ✓✓
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