Question

Four different integers form an increasing $A.P$. One of these numbers is equal to the sum of the squares of other three numbers. Find the sum of the numbers.

Answer 1

Let the numbers be $a-d,a,a+d,a+2d$
According to the hypothesis,
$(a-d{)}^2+a^2+(a+d{)}^2=a+2d$
i.e $2d^2-2d+3a^2-a=0$
$\therefore d=\frac{1}{2}[1\pm \sqrt{(1+2a-6a^2)}]$
Since, $d$ is a positive integer,
$1+2a-6a^2>0$
$a^2-\frac{a}{3}-\frac{1}{6}<0$
$(a-\frac{1-\sqrt{7}}{6})(a-\frac{1+\sqrt{7}}{6})<0$
$\left(\frac{1-\sqrt{7}}{6}\right)<a<\left(\frac{1+\sqrt{7}}{6}\right)$
Since $a$ is an integer,
$\therefore a=0$
then $d=\frac{1}{2}[1\pm 1]=1or0$. Since$d>0\therefore d=1$
Hence, the numbers are $-1,0,1,2$