Question

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

Answer 1

Since any odd positive integer n is of the form 4q+1 or 4q+3

If n=4q+1 then

n²=(4q+1)²=16q²+8q+1=8q(q+1)=8m+1

Where m=q(q+1)

If n=4q+3

Then n²=(4q+3)²=16q²+24q+9

=8(2q²+3q+1)+1=8m+1 where m=2q²+3q+1

Hence n² is of the form 8m+1

Answer 2

Answer:

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 , 8q +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

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