A polynomial $f(x)$ and its behavior as $f(t)>5$ Let $f(x)$ be a polynomial with integer coefficients. Suppose there are four distinct integers $p,q,r,s$ such that $f(p) = f(q) = f(r) = f(s) = 5$. If $t$ is an integer and $f(t)>5$, what is the smallest possible value of f(t)?
Could I get some hints?  This polynomial seems quite abstract, and is probably a quartic, but I can't account for multiple roots and such.  
 A: Let $g(x) =f(x) -5$. Then we know that $$g(x)=(x-p)(x-q)(x-r)(x-s)h(x)$$
for some polynomial $h(x)$. The condition $f(t)>5$ translates to $g(t)>0$.
Since $p,q,r,s,t$ are distinct integers, the smallest possible positive value of $(t-p)(t-q)(t-r)(t-s)$ is $4$ : the four numbers in the parentheses are all distinct integers $\neq0$, so the smallest value we can get from the product $(-2)\cdot(-1)\cdot1\cdot2$. }
The smallest possible positive value of $h(t)$ is $1$, since we must have $g(t)\neq 0$.
Thus the smallest possible value of $g(t)$ is $4$, and therefore the smallest possible value of $f(t)$ is $9$, and it is achieved for $t=2$ if we have
$$
f(x) =x(x-1)(x-3)(x-4) +5
$$
with $p,q,r,s$ being $0,1,3,4$ respectively.
A: Consider $g(x) = f(x)-5$. This polynomial will have zeroes at $p,q,r$, and $s$:
$$g(p)=g(q)=g(r)=g(s)=0$$
Note that $g(x) \neq 0$ because $g(t) > 0$. 

Now, as André Nicolas pointed out, there's a nonzero polynomial $h(x)$ (which must have integer coefficients) such that, for all $x$, 
$$g(x) = (x-p)(x-q)(x-r)(x-s)h(x)$$
If we wish to minimize the value of $f(t)$, we should minimize the value of $g(t)$, and to do this we must choose $h(x)=1$. 
So we have
$$g(t) = (t-p)(t-q)(t-r)(t-s)$$
How can we choose $p,q,r,s,t \in \mathbb{Z}$ to minimize this expression, given that it must be greater than $0$? We need the sign to be positive and the magnitude to be as small as possible. We'll want two of the factors to be $1$ and $-1$, and the other two to be $2$ and $-2$. This is clearly the minimum product.
We must finally show that it's possible to choose some $p,q,r,s,t$ to achieve this product of $4$. The following works:
$$p=2014 \qquad\, q=2015 \qquad\, r=2012 \qquad\, s=2011 \qquad\, t=2013$$

This means $g(t)$ has minimum $4$, so $f(t)$ has minimum $\boxed{9}$.
A: Hint: The polynomial has shape $f(x)=(x-p)(x-q)(x-r)(x-s)h(x)+5$.  Let $f(t)=5+b$. 
