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Consider the first number to be $x$ so that the consecutive number will be $ x + 1 $.

Now, according to the question, the sum of the squares $ x $ and $ x + 1 $ is 365. So, the equation is written as: $ {x^2} + {(x + 1)^2} = 365 $.

Solve the equation $ {x^2} + {(x + 1)^2} = 365 $ by using the splitting the middle term method to determine the value of $x$ as:

$

{x^2} + {(x + 1)^2} = 365 \\

{x^2} + {x^2} + 2x + 1 = 365 \\

2{x^2} + 2x + 1 - 365 = 0 \\

2{x^2} + 2x - 364 = 0 \\

{x^2} + x - 182 = 0 \\

$

Now, split the middle term into two parts such that the sum will be the coefficient of $ x $ and the product will be the product of the coefficient of $ {x^2} $ and $ {x^0} $.

$

{x^2} + 14x - 13x - 182 = 0 \\

x(x + 14) - 13(x + 14) = 0 \\

(x + 14)(x - 13) = 0 \\

x = 13, - 14 \\

$

As the question is asking for the positive consecutive numbers only, so drop $ x = - 14 $ and the desired value of $ x $ is 13.

Now, substitute $ x = 13 $ in the function $ x + 1 $ so as to determine the value of the second number as: $ x + 1 = 13 + 1 = 14 $.

Hence, 13 and 14 are two consecutive positive integers, the sum of whose square is 365.

1,2,3,4,… is a series of general, consecutive numbers.

2,4,6,8,… is a series of even consecutive numbers.

1,3,5,7,… is a series of odd consecutive numbers.