The product of four consecutive terms of an arithmetic progression of integers plus the fourth power of the common difference is always a perfect square. Verify this identity by incorporating symmetry into the notation.
I worked on this problem to solidify my heuristic for symmetry.
My solution is as follows:
$$(a_n-2d)(a_n-d)(a_n)(a_n+d)=a_n^4-2da_n^3-d^2a_n^2+2d^3a_n$$ Adding $d^4$ we have: $$a_n^4-2da_n^3-d^2a_n^2+2d^3a_n+d^4$$ I substituted $x=a_n$ and I obtained $$x^4-2dx^3-d^2x^2+2d^3x+d^4=(x^2-dx^2-d^2)^2$$
But I feel like I didn't use symmetry and I missed the point of the problem. What do you think?
Just let common difference be $2d$, then
$(a-3d)(a-d)(a+d)(a+3d)+16d^4= a^4-10d^2a^2+25d^4 = (a^2-5d^2)^2$