call the Polynomial f. f(x(i)) = +-1 for each i=1,…,7
Suppose f=g*h, two Polynomials of degree 1 at least.
so one of them is of degree 3 at most. WLOG it’s g.
since the polynomials are of integer coefficients, and x(i) are integers too, f(x(i))=g(x(i))*h(x(i)), and all g(x(i)) must divide +-1, so g(x(i)) = +-1 for each i = 1,…,7
so we now have g, a polynomial of degree at most 3, getting +-1 on 7 integers.
one of the values +1 or -1 is received 4 times at least. WLOG it’s 1
so we now have g-1, a polynomial of degree at most 3, getting 0 on 4 integers at least.
so g-1 == 0 every place.
contradiction (g’s degree is at least 1)

Thanks guys!

Rani,

You are welcome to still write down the answers to the Ugly and the Good. Although the Bad is much harder, it could be given in Discrete Math as well, as its proof relies entirely on the pigeonhole principle too.

What about the Unrelated?