How can we prove that $[ {c^n - (a^n+b^n)} + d^n ] / (ab-cd)$ will yield integers always? 
Let us say $a<b<c$ and $(a+b)=(c+d)$ as well as $n\geq2$. Then it is found for all values of $n$ that 
  $$
  \frac{{c^n-(a^n+b^n)}+d^n}{ab-cd} \in \mathbb{Z}
$$

Can you explain the reason for this?
 A: Here is a proof by induction:
For $n=0$ or $n=1$, we have that
$$\frac{c^n-(a^n+b^n)+d^n}{ab-cd}=0 \in\mathbb{Z}$$
Now suppose that for some $n\geq 2$, we have that
$$\frac{c^k-(a^k+b^k)+d^k}{ab-cd}$$
is an integer for all $k < n$. We will show that then
$$\frac{c^n-(a^n+b^n)+d^n}{ab-cd}$$
is also an integer.
The Binomial Theorem gives us that
$$(a+b)^n=a^n+b^n+\sum_{k=1}^{n-1} \binom{n}{k} a^kb^{n-k}$$
and
$$(c+d)^n=c^n+d^n+\sum_{k=1}^{n-1} \binom{n}{k} c^kd^{n-k}$$
so that
$$c^n-(a^n-b^n)+d^n=\sum_{k=1}^{n-1}\binom{n}{k}\left(c^kd^{n-k}-a^kb^{n-k}\right)$$
We want to show that this expression is congruent to $0$ modulo $ab-cd$.
Now
$$\sum_{k=1}^{n-1}\binom{n}{k}\left(c^kd^{n-k}-a^kb^{n-k}\right) \equiv \sum_{0<2k<n}\binom{n}{k}\left(c^kd^{n-k}+c^{n-k}d^k-a^kb^{n-k}-a^{n-k}b^k\right) \mod (ab-cd)$$
(The sum ranging only over $2k<n$ is still valid when $n$ is even, since if $n=2k$, then
$$c^kd^{n-k}-a^kb^{n-k}\equiv (ab)^k-(ab)^k\equiv 0 \mod (ab-cd)$$
and so we can exclude the $2k=n$ term)
But
$$\sum_{0<2k<n}\binom{n}{k}\left(c^kd^{n-k}+c^{n-k}d^k-a^kb^{n-k}-a^{n-k}b^k\right)=\sum_{0<2k<n}\binom{n}{k}\left(c^kd^k(c^{n-2k}+d^{n-2k})-a^kb^k(a^{n-2k}+b^{n-2k})\right)
\equiv \sum_{0<2k<n}\binom{n}{k}\left(a^kb^k((c^{n-2k}+d^{n-2k})-(a^{n-2k}+b^{n-2k}))\right) \mod (ab-cd)$$
Now by the inductive hypothesis,
$$(c^{n-2k}+d^{n-2k})-(a^{n-2k}+b^{n-2k})\equiv 0\mod (ab-cd)$$
for each $0<2k<n$, and so
$$\sum_{0<2k<n}\binom{n}{k}\left(a^kb^k((c^{n-2k}+d^{n-2k})-(a^{n-2k}+b^{n-2k}))\right) \equiv 0\mod (ab-cd)$$
and so
$$c^n-(a^n+b^n)+d^n\equiv 0 \mod (ab-cd)$$
i.e.
$$\frac{c^n-(a^n+b^n)+d^n}{ab-cd}\in\mathbb{Z}$$
for $n$ as well.
Thus, by the principle of mathematical induction, we have that
$$\frac{c^n-(a^n+b^n)+d^n}{ab-cd}\in\mathbb{Z}$$
for all natural numbers $n$.
A: Note that $d=a+b-c$. Consider the polynomial :
$$
P(a,b,X)=X^n-(a^n+b^n)+(a+b-X)^n
$$
This is a polynomial of degree $n$ in $X$ (or degree $n-1$ if $n$ is odd), and we have
$P(a)=P(b)=0$. Since $a\neq b$, we deduce that $(X-a)(X-b)$
divides $P$ in ${\mathbb Z}[a,b,X]$ : there is a polynomial
$Q\in{\mathbb Z}[a,b,X]$ such that $P(a,b,X)=(X-a)(X-b)Q(a,b,X)$.
Then $\frac{c^n-(a^n+b^n)+d^n}{ab-cd}=Q(a,b,c)\in {\mathbb Z}$.
A: First we give a simple inductive proof then we give a conceptual interpretation of the induction as a special case of the uniqueness theorem for recurrences (difference equations).
Hint $\ $ The inductive step of a simple inductive proof follows by subtracting these
$$\begin{align}  
c^{n+2}+d^{n+2} &= (c+d)(c^{n+1}+d^{n+1}) - cd(c^n+d^n)\\
a^{n+2}+b^{n+2} &= (a+b)(a^{n+1}+b^{n+1}) - ab(a^n+b^n)\end{align}\qquad$$
Indeed $\,{\rm mod}\ ab\!-\!cd\!:\ $ by induction $\,c^k\!+d^k\equiv a^k\!+b^k\,$ for $\,k = n,\,n\!+\!1\,$ so subtracting
$$c^{n+2}\!+\!d^{n+2}\!-\!a^{n+2}\!-\!b^{n+2}\equiv \underbrace{(c\!+\!d\!-\!a\!-\!b)}_{\large =\, 0}(a^{n+1}\!+b^{n+1}) + \underbrace{(ab\!-\!cd)}_{\large \equiv\, 0}(a^n\!+b^n)\,\equiv\, 0$$
Remark $\ $ More generally this yields $\ \gcd(c\!+\!d\!-\!a\!-\!b,ab\!-\!cd)\mid c^n\!+d^n\!-a^n\!-b^n$
Conceptually $\ f_n = a^n\!+b^n$ and $\,{\bar f}_{\!n} =  c^n+d^n$ are both solutions of the same recurrence $\,  f_{n+2} = A f_{n+1} - B f_n\,$ for $\, a+b\equiv A\equiv c+d,\ $ $\,ab \equiv B\equiv cd,\,$ with same initial conditions $\,f_0 \equiv 2,\ f_1 \equiv a+b\equiv c+d,\,$ thus $\,f_n\equiv {\bar f}_{\!n}\,$ for all $\,n\,$ by the same simple induction as above. This is just  the uniqueness theorem for recurrences. 
A: Sketch of a symmetrical functions proof: let's denote $s=a+b=c+d$, $p=ab$, $q=cd$.
By Newton'identities, we know $a^n+b^n$ is a polynomial $P_n(s,p)\in \mathbf Z[s,p]$ and $c^n+d^n=P_n(s,q)$. 
What we have to show is $P_n(s,p)-P_n(s,q)$  is divisible by $p-q$. This results from Taylor's formula for polynomials.
Details: Indeed, we know Taylor's  formula is an exact formula for polynomials. Let's consider $P_n(s,p)$ and $P_n(s,q)$ as the evaluation of a polynomial $P_n(s,x)$ at $p$ and $q$ respectively. If $d$ is its degree, Taylor's formula  can be written as:
$$P_n(s,p)-P_n(s,q)=(p-q)P'_n(s,q)+(p-q)^2\frac{P''_n(s,q)}{2}+\dots+(p-q)^d\frac{P^{(d)}_n(s,q)}{d\,!}$$
This proves $P_n(s,p)-P_n(s,q)$ is divisible by $p-q$, since all polynomials $\,\dfrac{P_n^{(k)}(s,x)}{k\,!}$ lis in $\mathbf Z[x]$.
A: High-school math proof:
We need to make use of the given condition. Let $u=a+b=c+d$, $v=a-b$ and $w=c-d$. Rewrite $a=(u+v)/2$, $b=(u-v)/2$, $c=(u+w)/2$ and $d=(u-w)/2$.
The denominator is then
$$ab-cd=\frac14\left(u^2-v^2-(u^2-w^2)\right)=\frac14\left(w^2-v^2\right)$$
The numerator is
$$\frac{1}{4^n}\left((u+w)^n+(u-w)^n-((u+v)^n+(u-v)^n)\right)$$
$$=\frac{1}{4^n}\left(2\sum_{k=0,2,4\ldots}^n {n \choose k} u^{n-k} w^k -2\sum_{k=0,2,4\ldots}^n {n \choose k} u^{n-k} v^k \right)$$
$$=\frac{2}{4^n}\sum_{k=0,2,4\ldots}^n {n \choose k} u^{n-k} (w^k-v^k) $$
Because $k=2m$ is even, you just need to show that $w^2-v^2$ divides $(w^2)^m-(v^2)^m$, which is solved by a high-school level equality $a^n-b^n=(a-b)(a^{n-1}+\cdots)$.
