# Sequence and Series ?

## The fourth power of the common difference of an arithmetic progression with integer entries is added to the product of any four consecutive terms of it. Prove that the resulting sum is the square of an integer.

Apr 6, 2017

Refer to The Explanation.

#### Explanation:

As we need to consider $4$ consecutive terms of an A.P., it will

be easier to take the

$4 \text{ terms } x - 3 y , x - y , x + y , \mathmr{and} , x + 3 y .$

The common difference, then, is $2 y .$

To prove that, $\left(x - 3 y\right) \left(x - y\right) \left(x + y\right) \left(x + 3 y\right) + {\left(2 y\right)}^{4}$ is a perfect suare.

$\text{The Exp.=} \left(x - 3 y\right) \left(x - y\right) \left(x + y\right) \left(x + 3 y\right) + {\left(2 y\right)}^{4} ,$

$= \left[\left\{\left(x - 3 y\right) \left(x + 3 y\right)\right\} \left\{\left(x - y\right) \left(x + y\right)\right\}\right] + 16 {y}^{4} ,$

$= \left[\left({x}^{2} - 9 {y}^{2}\right) \left({x}^{2} - {y}^{2}\right)\right] + 16 {y}^{4} ,$

$= {x}^{4} - 10 {x}^{2} {y}^{2} + 9 {y}^{4} + 16 {y}^{4} ,$

$= {x}^{4} - 10 {x}^{2} {y}^{2} + 25 {y}^{4} ,$

$= {\left({x}^{2} - 5 {y}^{2}\right)}^{2.}$

Hence, the Proof.