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 square of an integer.: Sequence & Series (Mathematics)

Let four consecutive terms of the A.P. be

a-3 d, a-d, a+d

and

a+3 d

. Common difference is

2 d

.\nGiven

a-3 d, a-d, a+d

and

a+3 d

are integers. Therefore,

2 d

is also an integer.\nNow,

E=(a-3 d)(a-d)(a+d)(a+3 d)+(2 d)^4

\n\n

\n\begin{aligned}\n& =\left(a^2-9 d^2\right)\left(a^2-d^2\right)+16 d^4 \\\n& =a^4-10 d^2 a^2+9 d^4+16 d^4\n\end{aligned}\n

E=\left(a^2-5 d^2\right)^2

is an integer\n(As

a-3 d, a+3 d

and

2 d

are integers

\Rightarrow a^2-5 d^2

is also an integer)\nThus,

E

is the square of an integer

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