Question

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

Answer 1

Let the consecutive numbers be $x,x+1$

Sum of their squares $x^2+(x+1{)}^2=365x^2+x^2+2x+1=3652x^2+2x-364=0x^2+x-182=0x^2+14x-13x-182=0x(x+14)-13(x+14)=0(x+14\left)\right(x-13)=0x=-14,13$

Since, x can't be a negative integers.
$\therefore x=13$
Hence, two consecutive positive integers$=13,14$