Question

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Question: 1)

Answer 1

Let the positive integer be a and let b be 4

According to Euclid's Division Lemma
a=bq+r
a=4m+r where 0≤r<b
so, r= 0,1,2,3
a=4m (even)
a=4m+1 (odd)
a=4m+2 (even)
a=4m+3 (odd)

Taking a=4m+1
Squaring both sides
a²= (4m+1)²
a²=16m²+1+8m
a²= 8(2m²+m)+1
a²=8q+1 where q=2m²+m

Taking a=4m+3
Squaring both sides
a²= (4m+3)²
a²=16m²+9+24m
a²= 16m²+24m+8+1
a²=8(2m²+3m+1)+1
a²=8q+1 where q=2m²+3m+1

Answer 2

Let a be any positive odd integer Which when divided by 4 gives m as quotient and r as remainder.


By Euclid's division lemma

a = bq + r


Where

0 ≥ r < b

we can conclude that

a = 4m + r

Where r = 0 , 1 , 2 , 3


a = 4m

a = 4m + 1

a = 4m + 2

a = 4m + 3


So,

a = 4m + 1 and a = 4m + 3 are odd

coz when we multiply any number by 4 we get even number as resultant and by adding odd in that resultant we get odd.


Case :- 1

a = 4m + 1

a² = ( 4m + 1 )²

( Using identity ( a + b )² = a² + b² + 2ab )

a² = 16m² + 1 + 8m

a² = 8q ( Where q = 2m² + m ) + 1

a² = 8q + 1


Case :- 2

a = 4m + 3

a² = ( 4m + 3 )²

a² = 16m² + 9 + 24m

a² = 16m² + 8 + 1 + 24m

a² = 8 q ( where q = 2m² + 1 + 3m ) + 1

a² = 8q + 1


Hence proved !!!