# Odd numbersFind two consecutive of the odd natural numbers in the sum of the whose squares is 202.

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Let the two odd numbers be 2n + 1 and 2n + 3

The sum of their squares is 202

=> (2n + 1)^2 + (2n + 3)^2 = 202

=> 4n^2 + 4n + 1 + 4n^2 + 12n + 9 = 202

=> 8n^2 + 16n - 192 = 0

=> n^2 + 2n - 24 = 0

=> n^2 + 6n - 4n - 24 = 0

=> n(n + 6) - 4(n + 6) = 0

=> (n - 4)(n + 6) = 0

=> n = 4 and n = -6

**This gives the pairs of odd numbers as (9, 11) and (-11, -9)**

The odd number is 2k + 1.

The other consecutive odd number is 2k + 3.

The sum of their squares is 202:

(2k+1)^2 + (2k+3)^2 = 202

We'll raise to square:

4k^2 + 4k + 1 + 4k^2 + 12k + 9 = 202

We'll combine like terms:

8k^2 + 16k - 192 = 0

We'll divide by 8:

k^2 + 2k - 24 = 0

We'll apply the quadratic formula:

k1 = [-1 + sqrt(4 + 96)]/2

k1 = (-1 + sqrt 100)/2

k1 = (-1+10)/2

k1 = 9/2

k2 = -11/2

The first odd number is:

2k + 1 = 2*9/2 = 9

The second odd number is 11.

Since the numbers are natural, the second solution of k is rejected.