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

Hint: We solve this problem by considering the required numbers as \['n'\] and \['n+1'\] 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 \['n'\] . 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 \['n'\] .
Since, the numbers are consecutive the other number can be get by adding ‘1’ to \['n'\] .
So, the two numbers are \['n'\] and \['n+1'\]
We are given that the sum of squares of these two numbers is 365.
Now, by converting this statement to mathematical equation we get
 \[\Rightarrow {{n}^{2}}+{{\left( n+1 \right)}^{2}}=365\]
We know that the formula \[{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\]
By using this formula to above equation we get
 \[\begin{align}
  & \Rightarrow {{n}^{2}}+\left( {{n}^{2}}+2n+1 \right)=365 \\
 & \Rightarrow 2{{n}^{2}}+2n-364=0 \\
\end{align}\]
Let us use the factorization method to solve this quadratic equation.
Here, we can write the middle term \['2n'\] as follows
 \[\Rightarrow 2{{n}^{2}}+28n-26n-364=0\]
Now, by taking the common terms in first two terms and common terms from last two terms we get
 \[\begin{align}
  & \Rightarrow 2n\left( n+14 \right)-26\left( n+14 \right)=0 \\
 & \Rightarrow \left( n+14 \right)\left( 2n-26 \right)=0 \\
 & \Rightarrow n=-14 or 13 \\
\end{align}\]
Here, we got two values for \['n'\] as ‘-14’ and ‘13’.
In the question, we are given to find positive integers.
So, ‘-14’ should not be taken.
Therefore the number = \['n'\] 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 \['n'\] as
 \[\Rightarrow {{n}^{2}}+{{\left( n+1 \right)}^{2}}=365\]
We know that the formula of roots of equation \[a{{x}^{2}}+bx+c=0\] is
 \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
By using this formula let us find value of \['n'\] as
 \[\begin{align}
  & \Rightarrow n=\dfrac{-2\pm \sqrt{{{2}^{2}}-4\left( 2 \right)\left( -364 \right)}}{2\left( 2 \right)} \\
 & \Rightarrow n=\dfrac{-2\pm \sqrt{4+2912}}{4} \\
 & \Rightarrow n=\dfrac{-2\pm 54}{4} \\
\end{align}\]
Here, one root is for \['+'\] and other for \['-'\] . By separating we get
 \[\begin{align}
  & \Rightarrow n=\dfrac{-2+54}{4}or\dfrac{-2-54}{4} \\
 & \Rightarrow n=13or-14 \\
\end{align}\]
Here, we got two values for \['n'\] as ‘-14’ and ‘13’.
In the question, we are given to find positive integers.
So, ‘-14’ should not be taken.
Therefore the number = \['n'\] 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.