Find two consecutive of the odd natural numbers in the sum of the whose squares is 202

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.