The 4th power of the common difference of an A.P. with integer entries is added to the product of any 4 consecutive terms of it. Prove that the resulting sum is the square of an integer.
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1708\hspace{-16}$Let $\bf{4}$ Terms of Arithmetic Progression be $\bf{a-3d\;,a-d\;,a+d\;,a+3d}$\\\\ So Common Difference is $\bf{=2d}$\\\\ Now Accroding To the Question\\\\ $\bf{(a-3d)\times (a-d)\times(a+d)\times(a+3d)+(2d)^4}$\\\\ $\bf{=(a^2-9d^2)\times (a^2-d^2)+16d^4}$\\\\ $\bf{=(a^2)^2-10a^2.d^2+9d^4+16d^4=a^4-10a^2.d^2+(5d^2)^2}$\\\\ $\bf{=\left(a^2-5d^2\right)^2=}$Perfect Square\\\\ Where $\bf{a\;,d\in\mathbb{Z}}$