Prove the following statement.
the square of any odd integer has the form 8m+1 for some integer m.

As per the question,

Let a be any positive integer and b = 4.

According to Euclid division lemma , a = 4q + r

where 0 ≤ r < b.

Thus,

r = 0, 1, 2, 3

Since, a is an odd integer, and

The only valid value of r = 1 and 3

So a = 4q + 1 or 4q + 3

Case 1 :- When a = 4q + 1

On squaring both sides, we get

a² = (4q + 1)²

   = 16q² + 8q + 1

   = 8(2q² + q) + 1

   = 8m + 1 , where m = 2q² + q

Case 2 :- when a = 4q + 3

On squaring both sides, we get

a² = (4q + 3)²

   = 16q² + 24q + 9

   = 8 (2q² + 3q + 1) + 1

   = 8m +1, where m = 2q² + 3q +1

Now,

We can see that at every odd values of r, square of a is in the form of 8m +1.

Also we know, a = 4q +1 and 4q +3 are not divisible by 2 means these all numbers are odd numbers.

Hence , it is clear that square of an odd positive is in form of 8m +1

its a little blurry tbh

step-by-step explanation:

step-by-step explanation:

y = | -2(-2) |     →     y = | 4 |     →     y = 4       ⇒   (-2, 4)

y = | -2(-1) |     →     y = | 2 |     →       y =   2       ⇒   (-1, 2)    

y = | -2(0) |     →     y = | 0 |     →       y = 0         ⇒   (0, 0)

y = | -2(1) |   c →     y = | -2 |     →     y =   2         ⇒   (1, 2)    

y = | -2(2) |     →     y = | - 4 |     →     y =   4         ⇒   (2, 4)