Is $\frac{a^4}{(b-a)(c-a)}+\frac{b^4}{(c-b)(a-b)}+\frac{c^4}{(a-c)(b-c)} $ always an integer? In a textbook I found the rather strange identity:
$$ \frac{2^4}{(5-2)(3-2)}+\frac{3^4}{(5-3)(3-2)}+\frac{5^4}{(5-3)(5-2)}= \frac{414}{6}=69 $$
just kind if out of nowhere and I wonder if it generalizes and why. Perhaps it is that:
$$ \frac{a^4}{(b-a)(c-a)}+\frac{b^4}{(c-b)(a-b)}+\frac{c^4}{(a-c)(b-c)}  \tag{$\ast$} $$
is always an integer?  On the same page I found the formula:
$$ \left|\begin{array}{cccc} 1  & a & a^2 & a^4 \\ 
1  & b & b^2 & b^4 \\ 
1  & c & c^2 & c^4 \\ 
1  & d & d^2 & d^4 \\ \end{array} \right| = P \times (a+b+c+d)$$
My guess is that $P$ is short-hand for the product of all the different pairs of numbers:
$$ P = (a-b)(a-c)(a-d)(b-c)(b-d)(c-d) $$
This is not the right formula I'm afraid.   However, I do see:
$$  \left|\begin{array}{cccc} 1  & 2 & 16  \\ 
1  & 3 & 81 \\ 
1  & 5 & 625 \end{array} \right| = P \times \stackrel{2}{C}(2,3,5)$$
which could mean anything.  The book is in German - which I do not understand, and the man has invented this wonderful symbol $\stackrel{2}{C}$.

Mainly I am wondering if other integer identities can be found this way.  Or is it only the case for $(a,b,c) = (2,3,5)$.

The statement is true even when you replace the numerators by any polynomial with integer coefficients or increase the number of variables. More precisely,

For any $f(x) \in \mathbb{Z}[x]$ and distinct $a_1, \ldots, a_n \in \mathbb{Z}$,
  $\displaystyle\;\sum\limits_{k=1}^n \frac{f(a_k)}{\prod\limits_{j=1,\ne k}^n (a_k - a_j)}\;$ is an integer.

I will only prove the statement for $3$ distinct integers $a,b,c$. The proof for other $n$ is similar.
For any $f(x) \in \mathbb{Z}[x]$, long divide it by $(x-a)(x-b)(x-c)$.
This gives us two polynomials $g(x), h(x) \in \mathbb{Z}[x]$ such that
$$f(x) = (x-a)(x-b)(x-c)g(x) + h(x)$$
and $h(x) = Ax^2 + Bx + C$ is at most quadratic.
Apply partial fraction decomposition to $\displaystyle\;\frac{h(x)}{(x-a)(x-b)(x-c)}\;$ and notice
$$f(a) = h(a),\quad f(b) = h(b)\quad\text{ and }\quad f(c) = h(c)$$
We obtain
$$\begin{align}
\frac{h(x)}{(x-a)(x-b)(x-c)} &=
\frac{1}{x-a}\frac{h(a)}{(a-b)(a-c)} + \frac{1}{x-b}\frac{h(b)}{(b-a)(b-c)}
+ \frac{1}{x-c}\frac{h(c)}{(c-a)(c-b)}\\
&= \frac{1}{x-a}\frac{f(a)}{(a-b)(a-c)} + \frac{1}{x-b}\frac{f(b)}{(b-a)(b-c)}
+ \frac{1}{x-c}\frac{f(c)}{(c-a)(c-b)}
\end{align}
$$
This implies
$$\frac{f(a)}{(a-b)(a-c)} + \frac{f(b)}{(b-a)(b-c)} + \frac{f(c)}{(c-a)(c-b)}
= \lim_{x\to\infty} \frac{xh(x)}{(x-a)(x-b)(x-c)} = A$$
The coefficients of $x^2$ in $h(x)$ which is an integer.
For the original problem, we can take $f(x) = x^4$ and conclude
$$\frac{a^4}{(a-b)(a-c)} + \frac{b^4}{(b-a)(b-c)} + \frac{c^4}{(c-a)(c-b)} \in \mathbb{Z}
\quad\text{ for distinct }\; a,b,c \in \mathbb{Z}
$$
