Question

The fourth power of a common difference $\left(d^4\right)$ of an $A.P.$ in which all terms are integer is added to product of any four consecutive terms of it. Then prove that it is a perfect square.

Answer 1

Let the terms in AP be $\alpha -3d,\alpha -d,\alpha +d,\alpha +3d$ such that

$a=\alpha$ , common difference $\left(d^{\prime }\right)=2d$

By conditions $(\alpha -3d)(\alpha +3d)(\alpha -d)(\alpha +d)+(2d{)}^4$

$=({\alpha }^2-9d^2)({\alpha }^2-d^2)+16d^4$

$={\alpha }^4-{\alpha }^2{d}^2-9{\alpha }^2{d}^2+9d^4+16d^4$

$={\alpha }^4-10{\alpha }^2{d}^2+25d^4$

$=({\alpha }^2-5d^2)$

Which is a perfect square

Hence, proved