Sure of any positive odd integer is of the form 8m+1
Answers
Any odd positive integer is of the form 4q + 1 or 4q + 3 for some integer q.
When n = 4q + 1,
⇒ n2 is in the form 8m + 1
Hence, n2 is in the form 8m + 1 if n is an odd positive integer .
Step-by-step explanation:
According to Euclid division lemma , a = bq + r where 0 ≤ r < b
Here we assume b = 8 and r ∈ [1, 7 ]
means r = 1, 2, 3, .....7
Then, a = 8q + r
Case 1 :- when r = 1 , a = 8q + 1
squaring both sides,
a² = (8q + 1)² = 64²q² + 16q + 1 = 8(8q² + 2q) + 1
= 8m + 1 , where m = 8q² + 2q
case 2 :- when r = 2 , a = 8q + 2
case 2 :- when r = 2 , a = 8q + 2 squaring both sides,
a² = (8q + 2)² = 64q² + 32q + 4 ≠ 8m +1
[ means when r is an even number it is not in the form of 8m + 1 ]
Case 3 :- when r = 3 , a = 8q + 3
Case 3 :- when r = 3 , a = 8q + 3 squaring both sides,
a² = (8q + 3)² = 64q² + 48q + 9 = 8(8q² + 6q + 1) + 1
= 8m + 1 , where m = 8q² + 6q + 1
You can see that at every odd values of r
square of a is in the form of 8m +1
But at every even Values of r square of a
isn't in the form of 8m +1 .
Also we know, a = 8q +1 , 8q +3 , 8q + 5 ,
9q +7 are not divisible by 2 means these
all numbers are odd numbers
Hence , it is clear that square of an odd
positive is in form of 8m +1