How to prove that $a.b.c.d + {( b - a )}^{4} $ is square of a natural number ( a , b ,c , d are four consecutive elements of an arithmetic sequence)

Question

How to prove that $a.b.c.d + {( b - a )}^{4} $ is square of a natural number ( $a , b ,c , d$ are four consecutive elements of an arithmetic sequence) , I have really no idea about this one , I tried to rewrite it as $ (c - 2d).(c - d).c.(c+d) + d^4 $ but it was useless.

Answer 1

Let $a = k-3p \\ b= k-p \\ c=k+p \\ d= k+3p $

then $$abcd + (b-a)^4 = (k^2-5p^2)^2$$

Answer 2

Suppose that $a_n=kn+r$ is the arithmetic progression, and write $a=kn+r$, $b=k(n+1)+r$, $c=k(n+2)+r$, $d=k(n+3)+r$. Then $$ abcd+(b-a)^4=(k^2n^2 + 3k^2n + k^2 + 2knr + 3kr + r^2)^2, $$ which is indeed a square.

Answer 3

Let the terms of the sequence be $a$, $a+k$, $a+2k$ and $a+3k$. Then, the expression we're looking at is equal to $$a(a+k)(a+2k)(a+3k)+k^4=a^4+6ka^3+11k^2a^2+6k^3a+k^4$$

If this is meant to be a square, it'd have to be square of some combination of $a^2$, $ak$ and $k^2$ (since our expression contains only terms of total degree $4$). Clearly, $a^2$ and $k^2$ must appear with coefficient $1$ or $-1$ and a short calculation shows that the right combination is $$a(a+k)(a+2k)(a+3k)+k^4 = (a^2+3ak+k^2)^2$$

Thus, the sum really is a square.