2022/044) If p, q, r, s are consecutive terms of an arithmetic sequence with common difference n how do you prove that $pqrs + n^4$ is a perfect square?

 Without loss of generality set n = 2x ( x need noy be integer so we can choose  

p = m - 3x, q = m - x, r = m + x, s = m + 3x

now $pqrs+n^2 = (m-3x)(m-x)(m+x)(m+3x)  + 16x^4$

$= (m- 3x)(m+ 3x)(m-x)(m+x) + 16x^4$ 

$= (m^2-9x^2)(m^2- x^2) + 16x^4$

$= m^4 - 10x^2m^2  + 9x^4 + 16x^4$

$=m^2 - 10x^2m + 25x^4 = (m^2 -5x^2)^2$ so it is a perfect square 

as it is integer square root is also integer