Question

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

Answer 1

Hint- The series of numbers that are more than 1 from its preceding number is known as consecutive numbers. The consecutive number series can be even as well as odd depending on the first term of the series and the difference between the numbers. If the difference between the two consecutive numbers is 2 and the starting term is an odd integer then, it is known as the series of the odd consecutive numbers, whereas if the difference between the two consecutive numbers is 2 and the starting term is an even integer, then it is known to be an even consecutive number series. For example, 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.

Here, the question is asking for the series of consecutive integers whose sum of each squared number is 365. So, consider the first term to be x, and then accordingly, the second term will be (x+1) and square both the numbers then find their sum.

Complete step by step solution:
Let us consider the two consecutive positive integers are $x,x + 1$
According to the question, the sum of the squares of the integers is 365.
So,
 ${x^2} + {\left( {x + 1} \right)^2} = 365$
Now, solving the above equation for the values of $x$:
$
  {x^2} + {x^2} + 1 + 2x = 365 \\
  2{x^2} + 2x + 1 - 365 = 0 \\
  2{x^2} + 2x - 364 = 0 \\
  {x^2} + x - 182 = 0 \\
 $

Now, splitting the middle term to solve the above quadratic equation:
$
  {x^2} + (14 - 13)x - 182 = 0 \\
  {x^2} + 14x - 13x - 182 = 0 \\
  x(x + 14) - 13(x + 14) = 0 \\
  (x + 14)(x - 13) = 0 \\
  x = - 14;13 \\
 $

As the question is asking for the two consecutive positive integers only, the value of $x = - 14$ should be dropped and $x = 13$ is to be considered as our result.

Now, substitute $x = 13$ in the function $x + 1$ to determine the second number we get,
$x + 1 = 13 + 1 = 14$

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

Note: It is interesting to note here that for a series of even or odd consecutive numbers, all the terms present in the series are even and odd, respectively.