Question

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

Answer 1

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
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
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

Answer 2

Answer:

ITS CORRECT, SIMPLE AND TIME FRIENDLY

Step-by-step explanation:

Any positive odd integer is of the form 2q + 1, where q is a whole number.

Therefore, (2q + 1)2

= 4q2 + 4q + 1 = 4q (q + 1) + 1,        ...(1)

q (q + 1) is either 0 or even. So, it is 2m, where m is a whole number.

Therefore, (2q + 1)2

= 4.2 m + 1

= 8 m + 1.

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