Let [i]Revolution[/i]$(x) = x^3 +Ux^2 +Sx + A$, where $U$, $S$, and $A$ are all integers and $U +S + A +1 = 1773$. Given that [i]Revolution[/i] has exactly two distinct nonzero integer roots $G$ and $B$, find the minimum value of $|GB|$.

[i]Proposed by Jacob Xu[/i]

[hide=Solution]
[i]Solution.[/i] $\boxed{392}$
Notice that $U + S + A + 1$ is just [i]Revolution[/i]$(1)$ so [i]Revolution[/i]$(1) = 1773$. Since $G$ and $B$ are integer roots we write [i]Revolution[/i]$(X) = (X-G)^2(X-B)$ without loss of generality. So Revolution$(1) = (1-G)^2(1-B) = 1773$. $1773$ can be factored as $32 \cdot 197$, so to minimize $|GB|$ we set $1-G = 3$ and $1-B = 197$. We get that $G = -2$ and $B = -196$ so $|GB| = \boxed{392}$.
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1. Given the polynomial *Revolution* $(x) = x^3 + Ux^2 + Sx + A$, we know that $U$, $S$, and $A$ are integers and that $U + S + A + 1 = 1773$.
2. Notice that $U + S + A + 1$ is just *Revolution* $(1)$, so we have:
   \[
   \text{Revolution}(1) = 1^3 + U \cdot 1^2 + S \cdot 1 + A = 1 + U + S + A = 1773
   \]
   Therefore, $\text{Revolution}(1) = 1773$.
3. Given that *Revolution* has exactly two distinct nonzero integer roots $G$ and $B$, we can write the polynomial as:
   \[
   \text{Revolution}(x) = (x - G)^2 (x - B)
   \]
   without loss of generality.
4. Substituting $x = 1$ into the polynomial, we get:
   \[
   \text{Revolution}(1) = (1 - G)^2 (1 - B) = 1773
   \]
5. To minimize $|GB|$, we need to factorize $1773$ and find appropriate values for $1 - G$ and $1 - B$.
6. The prime factorization of $1773$ is:
   \[
   1773 = 3 \times 591 = 3 \times 3 \times 197 = 3^2 \times 197
   \]
7. To minimize $|GB|$, we set $1 - G = 3$ and $1 - B = 197$. Solving for $G$ and $B$, we get:
   \[
   1 - G = 3 \implies G = 1 - 3 = -2
   \]
   \[
   1 - B = 197 \implies B = 1 - 197 = -196
   \]
8. Therefore, $|GB|$ is:
   \[
   |GB| = |-2 \times -196| = 392
   \]

The final answer is $\boxed{392}$.