Is there a polynomial $p(x)$ with integer coefficients, such that $p(a)=p(b)=p(c)=p(d)=4$ and $p(e) = 10$? I am trying to prove the following problem:

Prove that there's no such polynomial $p(x)$ with integer coefficients, such that $p(a) = p(b) = $ $p(c) = p(d) = 4$ and $p(e) = 10$, where $a, b, c, d, e$ are integers themselves and are distinct.

If $p(x)$ is the polynomial and $p(a) = p(b) = p(c) = p(d) = 4$ then it has the form $$p(x) = (x-a)(x-b)(x-c)(x-d)+4.$$
Now I can't figure out how I prove that there is no such integer $e$ that $p(e) = 10$.
I tried constructing various polynomials in Mathematica (as Lagrange Interpolating polynomial) and I always ended up having something like $$p(x) = \frac{(x-a)(x-b)(x-c)(x-d)}{\text{const}}+4$$ for a polynomial that interpolates points $(a,4), (b,4), (c,4), (d,4), (e,10)$.
I can't find a good argument that the polynomial $(x-a)(x-b)(x-c)(x-d)+4$ will always be divided by some $\text{const}$ to meet the $p(e)=10$ requirement, therefore there is no such polynomial with integer coefficients that.
Can anyone help?
 A: Hint $\ $ The key to proving that many problems of this type are unsolvable is to simply apply the Factor Theorem $\rm\ x-y\ |\ p(x)-p(y)\ $ in $\rm\:\mathbb Z[x,y]\ $ for $\rm\:p(x)\in \mathbb Z[x]\:.\:$ Specializing $\rm\ x,y = m,n\in\mathbb Z\:$ we deduce that $\rm\: m-n\ |\ p(m)-p(n)\:$ in $\rm\:\mathbb Z\:.\:$
For example, considering the specific example in Jyrki's answer, since $\rm\:p(a) = 4\:$ for $\rm\:a\ne 0\:$ we infer that $\rm\: a-0\ |\ p(a)-p(0) = 4-10\:,\:$ i.e. $\rm\: a\:|\:6\:$ for $\rm\:a\ne 0\:.\:$ These severe arithmetic constraints are enough to resolve the problem in many cases. A long time ago I once crafted a problem based on this combined with Pick's theorem that went unsolved for a long time till someone noticed the trick (it was John H. Conway if memory serves correct - it's not easy to pull the wool over his eyes!)
A: The problem is that in general the factors $(e-a), (e-b), (e-c), (e-d)$ are too large. But if we are careful, we can find examples, where this happens. For example, let $a=-1$, $b=-2$, $c=1$, $d=3$, $e=0$. Then
$$
p(x)=(x+1)(x+2)(x-1)(x-3)+4
$$
has the values $p(-1)=p(-2)=p(1)=p(3)=4$ and $p(0)=10$.
