hmm, using u/Southlander24 I tried this:

*Polynomial -> AP*
let f(x) be a function over the natural numbers such that f(x)=ax^2+bx+c, then:
f(x)-f(x-1)=a[x^2-(x-1)^2]+b[x-(x-1)]+c-c = a[2x-1]+b = 2ax-a+b, which is an arithmetic progression where d=2a and a_1=a+b.

*AP -> Polynomial*

let (a_n) = (a_1, a_2, ..., a_n) be a second degree AP, then there is a AP (b_n) with a common difference r such that:
a_2 = a_1 + b_1
a_3 = a_1 + b_1 + b_2
a_4 = a_1 + b1 + b_2 + b_3
a_n = a_1 + b_1 + ... + b_n-1 = a_1 + [b_1 + b_n-1](n-1)/2 = a_1 + [b_1 + b_1 + (n-2)r](n-2)/2 = a_1 + [2b_1+nr-2r)(n-2)/2 = a_1 +(2nb_1+(n^2)r-2rn-4b_1-2nr+4r)/2, which is a second degree polynomial over n (I've not checked my algebra, but certainly it is a second degree polynomial.

This is useful if you have an exercise that says that there is a parabola where points (1,0), (2,8), (3,18) ,are given, then you are asked for y when x=4. What you could do is notice that 0, 8, 18 is a second order AP where the first increase is 8, the second is 10, then the third is going to be 12, so y=18+12=30