Question

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

Hint:

For consecutive integers $n$ and $n+1$, sums of squares reduce to a simple quadratic: $n^2+(n+1)^2=2n^2+2n+1$.

Answer 1

Let the integers be $n$ and $n+1$.

Step 1: Form the equation.
\[ n^2+(n+1)^2=365 \ \Rightarrow\ 2n^2+2n+1=365 \] \[ \Rightarrow\ 2n^2+2n-364=0 \ \Rightarrow\ n^2+n-182=0. \]

Step 2: Solve the quadratic.
\[ \Delta=1+4\cdot 182=729 $\Rightarrow$ \sqrt{\Delta}=27. \] \[ n=\frac{-1\pm 27}{2}\ \Rightarrow\ n=13 \ \text{or}\ n=-14. \] Since $n$ is positive, $n=13$. Hence the integers are $13$ and $14$.
\boxed{13\ \text{and}\ 14}