Q4:

Find two consecutive positive integers, sum of whose squares is $365$.

Given: Two consecutive positive integers. The sum of the squares of these integers is $365$.

To Find: The two consecutive positive integers.

Step 1: Defining the Variables
Let the first positive integer be $x$.
Since the integers are consecutive, the next integer must be $x + 1$.
[Assumption: $x$ is a positive integer, so $x > 0$].

Step 2: Formulating the Equation
According to the problem statement, the sum of the squares of these two integers is $365$.
Mathematically, this is expressed as:
$x^2 + (x + 1)^2 = 365$

Step 3: Expanding and Simplifying the Equation
Apply the algebraic identity $(a + b)^2 = a^2 + 2ab + b^2$ to expand $(x + 1)^2$:
$x^2 + (x^2 + 2x + 1) = 365$
Combine like terms:
$2x^2 + 2x + 1 = 365$
Subtract $365$ from both sides to set the quadratic equation to the standard form $ax^2 + bx + c = 0$:
$2x^2 + 2x + 1 - 365 = 0$
$2x^2 + 2x - 364 = 0$

Step 4: Simplifying the Quadratic Equation
Divide the entire equation by $2$ to simplify the coefficients:
$x^2 + x - 182 = 0$

Step 5: Solving by Factorization
We need to find two numbers that multiply to $-182$ and add to $1$ (the coefficient of $x$).
The factors of $182$ are: $1 \times 182, 2 \times 91, 7 \times 26, 13 \times 14$.
Since the product is negative and the sum is positive, we choose $14$ and $-13$.
$x^2 + 14x - 13x - 182 = 0$
Group the terms:
$x(x + 14) - 13(x + 14) = 0$
Factor out the common binomial $(x + 14)$:
$(x + 14)(x - 13) = 0$

Step 6: Finding the Roots
Set each factor to zero:
1) $x + 14 = 0 \implies x = -14$
2) $x - 13 = 0 \implies x = 13$

Step 7: Evaluating the Results
The problem specifies that the integers must be positive.
Therefore, we reject $x = -14$.
We accept $x = 13$.

Step 8: Determining the Consecutive Integer
If the first integer is $x = 13$, then the second consecutive integer is:
$x + 1 = 13 + 1 = 14$.

Verification:
$13^2 + 14^2 = 169 + 196 = 365$. The condition is satisfied.

Final Answer: The two consecutive positive integers are 13 and 14.