# How do you find three consecutive integers such that the sum of twice the largest and the fourth power of the smallest is equal to the square of the remaining integer increased by 243?

May 21, 2016

Let the numbers be $x , \left(x + 1\right) , \left(x + 2\right)$

#### Explanation:

${x}^{4} + 2 \left(x + 2\right) = {\left(x + 1\right)}^{2} + 243$

${x}^{4} + 2 x + 4 = {x}^{2} + 2 x + 1 + 243$

${x}^{4} - {x}^{2} - 240 = 0$

$\left({x}^{2} - 16\right) \left({x}^{2} + 15\right) = 0$

${x}^{2} - 16 = 0 \mathmr{and} {x}^{2} + 15 = 0$

$x = \pm 4 \mathmr{and} x = \emptyset$

Therefore the numbers are $4 , 5 , 6$

Hopefully this helps.