Question

show that the square of any positive integer is either of the form 4q or 4q + 1 for some integer q


The person who will give right answer will reward 30point and will choose as brainlist.

Answer 1

Solution:

Let a be the positive integer and b = 4.

Then, by Euclid’s algorithm, a = 4m + r for

some integer m ≥ 0 and r = 0, 1, 2, 3 because 0 ≤ r < 4.

So, a = 4m or 4m + 1 or 4m + 2 or 4m + 3.

So, (4m)2 = 16m2 = 4(4m2)

= 4q, where q is some integer.

(4m + 1)2 = 16m2 + 8m + 1

= 4(4m2 + 2m) + 1

= 4q + 1, where q is some integer.

(4m + 2)2 = 16m2 + 16m + 4

= 4(4m2 + 4m + 1)

= 4q, where q is some integer.

(4m + 3)2 = 16m2 + 24m + 9

= 4(4m2 + 6m + 2) + 1

= 4q + 1, where q is some integer.

Hence, The square of any positive integer is either of the form 4q or 4q + 1, where q is some integer.

Answer 2

$\bold{Answer:-}$



$\bold{Step\: by \:step \:explanation}$





Using

$(a + b) {}^{2} = a {}^{2} + b {}^{2} + 2ab$






⏩⏩Refer to the attachment above for your solution ⬆ ⬆










Hope it helps!