Question

Find two consecutive positive integers, sum of whose squares is 613.

• A. 17 and 18
• B. 20 and 22
• C. 16 and 17
• D. 15 and 18

Answer 1

The correct option is A 17 and 18

Let us assume $x,x+1$ be the two consecutive positive integers, Therefore

$x^2+(x+1{)}^2=613$

$x^2+x^2+2x+1=613$

$2x^2+2x+1=613$

$2x^2+2x-612=0$

$2(x^2+x-306)=0$

$x^2+x-306=0$

$x^2-18x+17x-306=0$

$x(x-18)+17(x-18)=0$

$(x-18)(x+17)=0$

Therefore, $x=18$ or $-17$

Answer is $x=18$ and $17$