Question

The person who will be the first to answer my question Will chose as brainlist.

Answer 1

Here is your answer.

Let a be any positive integer and b = 4

Then, by Euclid''s algorithm a = 4q + r for some integer q 0 and 0 r < 4

Since, r = 0, 1, 2, 3

Therefore, a = 4q or 4q + 1 or 4q + 2 or 4q + 3

Since, a is an odd integer, o a = 4q + 1 or 4q + 3

Case I: When a = 4q + 1

Squaring both sides, we have,

a2 = (4q + 1)2

a2 = 16q2 + 1 + 8q

= 4(4q2 + 2q) + 1

= 4m + 1 where m = 4q2 + 2q

Case II: When a = 4q + 3

Squaring both sides, we have,

a2 = (4q +3)2

= 16q2 + 9 + 24q

= 16 q2 + 24q + 8 + 1

= 4(4q2 + 6q + 2) +1

= 4m +1 where m = 4q2 +7q + 2

Hence, a is of the form 4m + 1 for some integer m.

Answer 2

Hey friend

=>Let a be any odd integer.

we know that any odd integer
is of the form of 2q+1.

=>Then a=2q+1 for some integer q.

=>x²=(2q+1)²
=>x²=4q²+4q+1=4q+1,for some integer
m=q²+q.

So, the square of any odd integer is of the form of 4m+1.

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