Question

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

Answer 1

Find the required consecutive positive integers

Given that sum of squares of two consecutive positive integers is $365$

Let two consecutive numbers be $\mathrm{m}$ and $\mathrm{m}+1$.

So, according to question, we get;

${\mathrm{m}}^2+{\left(\mathrm{m}+1\right)}^2=365$

$\Rightarrow$ ${\mathrm{m}}^2+{\mathrm{m}}^2+2\mathrm{m}+1^2=365$ $\left[\because {\left(\mathrm{a}+\mathrm{b}\right)}^2={\mathrm{a}}^2+2\mathrm{ab}+{\mathrm{b}}^2\right]$

$\Rightarrow$ $2{\mathrm{m}}^2+2\mathrm{m}+1=365$

$\Rightarrow$ $2{\mathrm{m}}^2+2\mathrm{m}-364=0$

$\Rightarrow$ ${\mathrm{m}}^2+\mathrm{m}-182=0$

$\Rightarrow$ ${\mathrm{m}}^2+14\mathrm{m}-13\mathrm{m}-182=0$

$\Rightarrow$$\mathrm{m}\left(\mathrm{m}+14\right)-13\left(\mathrm{m}+14\right)=0$

$\Rightarrow$ $\left(\mathrm{m}+14\right)\left(\mathrm{m}-13\right)=0$

$\Rightarrow$ $\mathrm{m}=13,-14$

As, we need to find two positive integers so, $-14$ is neglected.

So, $m=13$ and $\mathrm{m}+1=14$

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