Question

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

Answer 1

Hint: We solve this problem by considering the required numbers as $${}^{\prime }{n}^{\prime }$$ and $${}^{\prime }n+1^{\prime }$$ because the required numbers should be consecutive. Then we use the given condition that is sum of squares of those two numbers as 365 to solve for $${}^{\prime }{n}^{\prime }$$ . So, we can get both the required numbers.

Complete step by step answer:
We are asked to find the two consecutive numbers.
Let us assume that one number is $${}^{\prime }{n}^{\prime }$$ .
Since, the numbers are consecutive the other number can be get by adding ‘1’ to $${}^{\prime }{n}^{\prime }$$ .
So, the two numbers are $${}^{\prime }{n}^{\prime }$$ and $${}^{\prime }n+1^{\prime }$$
We are given that the sum of squares of these two numbers is 365.
Now, by converting this statement to mathematical equation we get
 $$\Rightarrow n^2+{(n+1)}^2=365$$
We know that the formula $${(a+b)}^2=a^2+2ab+b^2$$
By using this formula to above equation we get
 $$\begin{array}{rl}& \Rightarrow n^2+(n^2+2n+1)=365\\ & \Rightarrow 2n^2+2n-364=0\end{array}$$
Let us use the factorization method to solve this quadratic equation.
Here, we can write the middle term $${}^{\prime }2n^{\prime }$$ as follows
 $$\Rightarrow 2n^2+28n-26n-364=0$$
Now, by taking the common terms in first two terms and common terms from last two terms we get
 $$\begin{array}{rl}& \Rightarrow 2n(n+14)-26(n+14)=0\\ & \Rightarrow (n+14)(2n-26)=0\\ & \Rightarrow n=-14or13\end{array}$$
Here, we got two values for $${}^{\prime }{n}^{\prime }$$ as ‘-14’ and ‘13’.
In the question, we are given to find positive integers.
So, ‘-14’ should not be taken.
Therefore the number = $${}^{\prime }{n}^{\prime }$$ is ‘13’.
If one number is 13 then another number is 14.

So, the consecutive numbers whose sum of squares is 365 are 13, 14.

Note: This problem is solved in another method.
We got the equation of $${}^{\prime }{n}^{\prime }$$ as
 $$\Rightarrow n^2+{(n+1)}^2=365$$
We know that the formula of roots of equation $$a{x}^2+bx+c=0$$ is
 $$x={\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}}$$
By using this formula let us find value of $${}^{\prime }{n}^{\prime }$$ as
 $$\begin{array}{rl}& \Rightarrow n={\displaystyle \frac{-2\pm \sqrt{2^2-4\left(2\right)(-364)}}{2\left(2\right)}}\\ & \Rightarrow n={\displaystyle \frac{-2\pm \sqrt{4+2912}}{4}}\\ & \Rightarrow n={\displaystyle \frac{-2\pm 54}{4}}\end{array}$$
Here, one root is for $${}^{\prime }{+}^{\prime }$$ and other for $${}^{\prime }{-}^{\prime }$$ . By separating we get
 $$\begin{array}{rl}& \Rightarrow n={\displaystyle \frac{-2+54}{4}}or{\displaystyle \frac{-2-54}{4}}\\ & \Rightarrow n=13or-14\end{array}$$
Here, we got two values for $${}^{\prime }{n}^{\prime }$$ as ‘-14’ and ‘13’.
In the question, we are given to find positive integers.
So, ‘-14’ should not be taken.
Therefore the number = $${}^{\prime }{n}^{\prime }$$ is ‘13’.
If one number is 13 then another number is adding 1 to it.
Therefore, the second number is ‘14’.
So, the consecutive numbers whose sum of squares is 365 are 13, 14.