# The fourth power of the common difference of an arithmetic progression is 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?

Jul 15, 2018

Let the common difference of an AP of integers be $2 d$.

Any four consecutive terms of the progression may be represented as $a - 3 d , a - d , a + d \mathmr{and} a + 3 d$ , where $a$ is an integer.

So the sum of the products of these four terms and fourth power of the common difference ${\left(2 d\right)}^{4}$ will be

$= \textcolor{b l u e}{\left(a - 3 d\right) \left(a - d\right) \left(a + d\right) \left(a + 3 d\right)} + \textcolor{red}{{\left(2 d\right)}^{4}}$

$= \textcolor{b l u e}{\left({a}^{2} - 9 {d}^{2}\right) \left({a}^{2} - {d}^{2}\right)} + \textcolor{red}{16 {d}^{4}}$

=color(blue)((a^4-10d^2a^2+9d^4)+color(red)(16d^4)

=color(green)((a^4-10d^2a^2+25d^4)

=color(green)((a^2-5d^2)^2 , which is a perfect square.