Question

Find two consecutive positive integers such that the sum of their squares is 841

Answer 1

Let x = 1st positive integer

Let x + 1 = 2nd positive integer

As give, their sum of squares is 841

$x^{2}+(x+1)^{2}=841$

$x^{2}+x^{2}+2x+1=841$

$2x^{2}+2x+1=841$

$2x^{2}+2x-840=0$

factor 2 out, $x^{2}+x-420=0$

$x^{2}+21x-20x-420=0$

$x(x+21)-20(x+21)=0$

(x+21)(x-20)

This gives, x=20 and =-21 (neglect the negative value)

So, x = 20

x+1 = 21

So, the two numbers are : 20 and 21