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 $(a,b,c) = (2,3,5)$.
 A: Because
$$ \frac{a^4}{(b-a)(c-a)}+\frac{b^4}{(c-b)(a-b)}+\frac{c^4}{(a-c)(b-c)}  = a^2 + a b + a c + b^2 + b c + c^2  \text{,}  $$
the result is certainly an integer for distinct integers $a$,$b$,$c$.  (It's undefined if any pair of them are equal.)
Your guess about $P$ being the product of differences of alphabetical pairs is correct.
Up to a minus sign, $\overset{2}{C}(2,3,5) = -69 $, is the negative of the first example you gave.
A: 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
$$\frac{a^4}{(a-b)(a-c)} + \frac{b^4}{(b-a)(b-c)} + \frac{c^4}{(c-a)(c-b)} \in \mathbb{Z}
\quad\text{ for distinct }\; a,b,c \in \mathbb{Z}
$$
A: According to Wolfram Alpha the expression 
$$\frac{a^4}{(b-a)(c-a)}+\frac{b^4}{(c-b)(b-a)}+\frac{c^4}{(c-a)(c-b)}  \tag{$\ast$}$$ is not always an integer (see https://www.wolframalpha.com/input/?i=%5Cfrac%7Ba%5E4%7D%7B(b-a)(c-a)%7D%2B%5Cfrac%7Bb%5E4%7D%7B(c-b)(b-a)%7D%2B%5Cfrac%7Bc%5E4%7D%7B(c-a)(c-b)%7D,+a%3D3,b%3D5,c%3D7). But the expression you wrote
$$\frac{a^4}{(b-a)(c-a)}+\frac{b^4}{(c-b)(a-b)}+\frac{c^4}{(a-c)(b-c)}  \tag{$\ast$}$$ is always an integer. (Note the difference with your example. The denominator of the second fraction changes sign).
With respect to the determinant note that:
\begin{align}\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| & \\ &\underbrace{=}_{(1)}\left|\begin{array}{cccc} 1  & a & a^2 & a^4 \\ 
0  & b-a & b^2-a^2 & b^4-a^4 \\ 
0  & c-b & c^2-b^2 & c^4-b^4 \\ 
0  & d-c & d^2-c^2 & d^4-c^4 \\ \end{array} \right| \\ &\underbrace{=}_{(2)}\left|\begin{array}{ccc}
b-a & b^2-a^2 & b^4-a^4 \\ 
c-b & c^2-b^2 & c^4-b^4 \\ 
d-c & d^2-c^2 & d^4-c^4 \\ \end{array} \right|\\ &\underbrace{=}_{(3)}(b-a)(c-b)(d-c)\left|\begin{array}{ccc}
1 & b+a & b^3+b^2a+ba^2+a^3 \\ 
1 & c+b & c^3+c^2b+cb^2+b^3 \\ 
1 & d+c & d^3+d^2c+dc^2+c^3 \\ \end{array}\right| \end{align}
where in $(1)$ we have made $F_i-F_1\to F_i, i=2,3,4,$ in $(2)$ we have used the Laplacian's expansion using the first column and in $(3)$ we get out of the determinand $b-a$ from the first row, $c-b$ from the second one and $d-c$ from the third one.
Now, proceeding in a similar way we have
\begin{align}\left|\begin{array}{ccc}
1 & b+a & b^3+b^2a+ba^2+a^3 \\ 
1 & c+b & c^3+c^2b+cb^2+b^3 \\ 
1 & d+c & d^3+d^2c+dc^2+c^3 \\ \end{array}\right| &\\&=\left|\begin{array}{ccc}
1 & b+a & b^3+b^2a+ba^2+a^3 \\ 
0 & c-a & c^3+c^2b+cb^2-b^2a-ba^2-a^3 \\ 
0 & d-b & d^3+d^2c+dc^2-c^2b-cb^2-b^3\\ \end{array}\right|\\&=\left|\begin{array}{ccc}
 c-a & c^3+c^2b+cb^2-b^2a-ba^2-a^3 \\ 
 d-b & d^3+d^2c+dc^2-c^2b-cb^2-b^3\\ \end{array}\right|\\&=(d-b)(d-a)(c-a)(a+b+c+d).\end{align}
So, your guess about $P$ is correct.
