Define $F(x)=x\cdot(x-1)\cdot(x-2)\cdot(x-3)$. Let $y,T$ be positive integers. Is it then true that $y=7$, $T=8$ is the only solution to this equation? $$\frac{F(y)}{F(T)}=\frac{1}{2}$$
Motivation for the question:
The motivation for the question comes from this exercise:
Assume you have a bag with green and red balls. You pick 4 balls at random, all of them are green. You are told that the probability for this to happen is $0.5$. How many green balls were there in the bag, and how many balls were there in total?
By trial and error it can be shown that $7$ green balls and $1$ red ball, 8 in total will solve this probability exercise. But I am wondering if this is the only possibility? Or are there other combinations of green and red balls that would work? If the answer to the original question is yes, then only green = 7, total = 8 would work.
Is there a way to show that no other possibility would work?
PS: I don't know much about number theory, those tags was edited in. But I understand the answer may lie in that category of mathematics.
Here is a hint: Note that $$ F(n)+1=n(n+1)(n+2)(n+3) + 1 = (n^2 + 3n + 1)^2. $$ So the equation $2F(n)=F(m)$ is equivalent to $$ 2(n^2+3n+1)^2-(m^2+3m+1)^2=1. $$ This is a (negative) Pell equation. From its solutions we can deduce the solutions of the original equation.
Edit: On needs to rewrite the four consecutive numbers in the product, i.e., $n=x-3$.