Question

Two consecutive positive integers sum of whose squares is $365$ are

Answer 1

,Finding the two consecutive positive integer:

Let the first integer be $x.$

The next consecutive positive integer will be $x+1.$

According to the given question,

$$\begin{gathered} x^2+(x+1{)}^2=365 \\ {x}^2+{(x+1)}^2=365 \end{gathered}$$

$x^2+(x^2+2x+1)=365$ $[:{(a+b)}^2=a^2+2ab+b^2]$

$$\begin{gathered} 2x^2+2x+1=365 \\ 2x^2+2x+1-365=0 \\ 2x^2+2x-364=0 \end{gathered}$$

$\begin{array}{rcl}2({\mathrm{x}}^2+\mathrm{x}-182)& =& 0\\ {\mathrm{x}}^2+\mathrm{x}-182& =& 0\\ {\mathrm{x}}^2+14\mathrm{x}-13\mathrm{x}-182& =& 0\\ \mathrm{x}(\mathrm{x}+14)-13(\mathrm{x}+14)& =& 0\\ (\mathrm{x}-13)(\mathrm{x}+14)& =& 0\\ \mathrm{x}-13& =& 0,x+14=0\\ x& =& 13,x=-14\end{array}$

The value of$x$ cannot be negative (because it is given that the integers are positive).

Therefore, the integer will be $x=13,x+1=14$