<s>[INST] 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 $(a,b,c) = (2,3,5)$.
 [/INST] 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.
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