by bibi

(city)

The sum of squares of two consecutive odd positive integers is 202. Find the integers.**Solution**

Let 2x + 1 be the first odd integer

Then, 2x + 3 will be the second integer

Now square each odd integer

(2x + 1)^{2} = (2x + 1) × (2x + 1)

(2x + 1) × (2x + 1) = 2x × 2x + 2x × 1 + 1 × 2x + 1 × 1

(2x + 1) × (2x + 1) = 4x^{2} + 2x + 2x + 1 = 4x^{2} + 4x + 1

(2x + 3)^{2} = (2x + 3) × (2x + 3)

(2x + 3) × (2x + 3) = 2x × 2x + 2x × 3 + 3 × 2x + 3 × 3

(2x + 3) × (2x + 3) = 4x^{2} + 6x + 6x + 9 = 4x^{2} + 12x + 9

Since the sum of the squares equal 202, we get

4x^{2} + 4x + 1 + 4x^{2} + 12x + 9 = 202

8x^{2} + 16x + 10 = 202

8x^{2} + 16x + 10 - 202 = 202 - 202

8x^{2} + 16x - 192 = 0

Divide everything by 8

We get x^{2} + 2x - 24 = 0

To factor the left side of this equation, look for factors of -24 that will add up to 2.

6 × -4 = -24 and 6 + -4 = 2.

We get (x + 6) × (x + -4) = 0

We have two equation to solve

x + 6 = 0 and x + -4 = 0

x = -6 and x = 4

x = -6 cannot be used since the integers are positive

we can use x = 4.

The first integer is 2x + 1 = 2 × 4 + 1 = 9

The seond integer is 2x + 3 = 2 × 4 + 3 = 11

Indeed 9^{2} + 11^{2} = 81 + 121 = 202