Question

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.

Answer 1

Any four consecutive integers in A.P. be taken as $(a-3d),(a-d),(a+d),(a+3d)$ so that common difference is $2d$ and their product is $(a^2-9d^2)(a^2-d^2)$
We have to evaluate
$(2d{)}^4+(a^2-9d^2\left)\right(a^2-d^2)$
or $a^4-10a^2{d}^2+25d^4=(a^2-5d^2{)}^2$
Since a and d are given to be integers therefore $(a^2-5d^2{)}^2$ is also an integer.