Question

The sum of the square of three consecutive natural numbers is 149. Find the numbers.

Answer 1

Let the three consecutive numbers be $x,x+1\text{and}x+2$

So,

According to given question,

$x^2+{(x+1)}^2+{(x+2)}^2=149$

$\Rightarrow x^2+x^2+2x+1+x^2+4x+4=149$

$\Rightarrow 3x^2+6x=149-5$

$\Rightarrow 3x^2+6x=144$

$\Rightarrow 3(x^2+2x)=144$

$\Rightarrow x^2+2x=48$

$\Rightarrow x^2+2x-48=0$

$\Rightarrow x^2+8x-6x-48=0$

$\Rightarrow x(x+8)-6(x+8)=0$

$\Rightarrow (x+8)(x-6)=0$

If $x+8=0,\Rightarrow x=-8$

If $x-6=0,\Rightarrow x=6$

Given that, the three consecutive numbers are natural numbers

$\therefore x\text{cannot be}-8$

Hence, $x,x+1,\text{and}x+2\Rightarrow 6,6+1\text{and}6+2$

$\Rightarrow 6,7\text{and}8$

Hence, the three consecutive numbers are $6,7\text{and}8$