Square of any positive odd integer is of the form 8m +1.
Answers
Solution :-
We know that
Euclid's Division Algorithm: Given positive integers a and b there exists unique pair of integers q and r satisfying a = bq+r , 0≤ r< b
Let a be any positive integer.
Let us apply Euclid's Division Algorithm with a and b = 8
a = 8q +r ---------(1)
Where, 0≤r<8
The possible values of r = 0,1,2,3,4,5,6,7.
Case -1:-
r = 0:
If r = 0 then (1) becomes
a = 8q+0
=> a = 8q
On squaring both sides then
=> a² = (8q)²
=> a² = 8×8q²
=> a² = 8m ------------(2)
Where, m = 8q²
Case-2:-
r = 1 :
If r = 1 then (1) becomes
a = 8q+1
On squaring both sides then
=> a² = (8q+1)²
=> a² = (8q)²+2(8q)(1)+1²
=> a² = 64q²+16q+1
=> a² = 8(8q²+2q)+1
=> a² = 8m+1 ------------(3)
Where, m = 8q²+2q
Case-3:-
r = 2:
If r = 2 then (1) becomes
a = 8q+2
On squaring both sides then
=> a² = (8q+2)²
=> a² = (8q)²+2(8q)(2)+2²
=> a² = 64q²+32q+4
=> a² = 64q²+32q+8-4
=> a² = 8(8q²+4q+1)-4
=> a² = 8m-4 ------------(4)
Where, m = 8q²+4q+1
Case-4:-
r = 3:
If r = 3 then (1) becomes
a = 8q+3
On squaring both sides then
=> a² = (8q+3)²
=> a² = (8q)²+2(8q)(3)+3²
=> a² = 64q²+48q+9
=> a² = 64q²+48q+8+1
=> a² = 8(8q²+6q+1)+1
=> a² = 8m+1 ------------(5)
Where, m = 8q²+6q+1
Case-5:-
r = 4:-
If r = 4 then (1) becomes
a = 8q+4
On squaring both sides then
=> a² = (8q+4)²
=> a² = (8q)²+2(8q)(4)+4²
=> a² = 64q²+64q+16
=> a² = 8(8q²+8q+2)
=> a² = 8m------------(6)
Where, m = 8q²+8q+2
Case-6:-
r = 5:-
If r = 5 then (1) becomes
a = 8q+5
On squaring both sides then
=> a² = (8q+5)²
=> a² = (8q)²+2(8q)(5)+5²
=> a² = 64q²+80q+25
=> a² = 64q²+80q+24+1
=> a² = 8(8q²+10q+3)+1
=> a² = 8m+1 ------------(7)
Where, m = 8q²+10q+3
Case-6:-
r = 6:-
If r = 6 then (1) becomes
a = 8q+6
On squaring both sides then
=> a² = (8q+6)²
=> a² = (8q)²+2(8q)(6)+6²
=> a² = 64q²+96q+36
=> a² = 64q²+96q+32+4
=> a² = 8(8q²+12q+4)+4
=> a² = 8m+4 ------------(8)
Where, m = 8q²+12q+4
Case-7:-
r = 7:-
If r = 7 then (1) becomes
a = 8q+7
On squaring both sides then
=> a² = (8q+7)²
=> a² = (8q)²+2(8q)(7)+7²
=> a² = 64q²+112q+49
=> a² = 64q²+112q+48+1
=> a² = 8(8q²+14q+6)+1
=> a² = 8m+1 ------------(9)
Where, m = 8q²+14q+6
From (3),(5),(7)&(9)
We conclude that " The square of any positive odd integer is of the form of 8m+1 ".
Hence, Proved.