Question

plzz solve this
fast​

Answer 1

Let a be any positive integer and b = 4.

By Euclid's Division Lemma:

a = 4q + r for some integer q ≥ 0 and 0 ≤ r < 4

Thus, r = 0, 1, 2, 3

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

Case 1:

When a = 4q + 1

Squaring both sides, we have, a² = (4q + 1)²

a² = 16q² + 1 + 8q

   = 8(2q² + q) + 1

    = 8m + 1 (where m = 2q² + q)

Case II:

When a = 4q + 3

Squaring both sides, we have,

a² = (4q +3)²

   = 16q² + 9 + 24q

   = 16 q² + 24q + 8 + 1

    = 8(2q² + 3q + 1) +1

    = 8m +1 (where m = 2q² + 3q + 1)

Hence, a is of the form 8m + 1 for some integer m

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Answer 2

Answer:

see above................