The key is to use strong induction First notice that $a-b+c-d | (a-b+c-d)(a+b+c+d) \implies a-b+c-d | ac-bd$ So we get $a-b+c-d | (ac)^k-(bd)^k$ Let $O_{n,k} = a^{k}c^{n-k}+a^{n-k}c^{k} -b^{n-k}d^k-b^kd^{n-k}$ Now we prove that if $a-b+c-d| a^{n-1-2i}-b^{n-1-2i}+c^{n-1-2i}-d^{n-1-2i}$ then $a-b+c-d$ divides a expression of the form \[O_{n-1,i} = a^ic^{n-1-i}+a^{n-1-i}c^i-b^id^{n-1-i}+b^{n-1-i}d^i\] Just notice \[O_{n-1,i}= a^ic^i(a^{n-1-2i}+c^{n-1-2i})-b^id^i(b^{n-1-2i}+d^{n-1-21})\]. The last expression is congruent to $a^ic^i(a^{n-1-2i}-b^{n-1-2i}+c^{n-1-2i}-d^{n-1-21}) \pmod {a-b+c-d}$ since $(ac)^i \equiv (bd)^i \pmod {a-b+c-d}$ and we are done. So lets use strong induction. Suppose that $a-b+c-d| a^i-b^i+c^i-d^i$ for every $i \le n-1$. If $n=2k$ notice that $a-b+c-d | (a^{k}-b^{k}+c^{k}-d^{k})(a^{k}+b^{k}+c^{k}+d^{k})$. The last implies that $a-b+c-d | a^{2k}-b^{2k}+c^{2k}-d^{2k} +2(a^kc^k-b^kd^k)$ and we are done. For the case $n$ odd just notice that $a^n-b^n+c^n-d^n = (a+c)(a^{n-1}-...+c^{n-1})-(b+d)(b^{n-1}+...+d^{n-1})$ Since $a+c \equiv b+d \pmod {a-b+c-d}$ it is enough to show \[a-b+c-d |a^{n-1}-...+c^{n-1} - (b^{n-1}+...+d^{n-1})\] However the last expression is of the form $a^{n-1}-b^{n-1}+c^{n-1}-d^{n-1} + \sum {O_{n-1,k}}$ and is divisible by $a-b+c-d$ so we are done.

Lemma. If $x-y \mid z_1-t_1$ and $x-y \mid z_2-t_2$ then $x-y \mid z_1z_2 - t_1t_2$. Proof. We have that $z_i \equiv t_i \pmod{x-y}$ for $i=1,2$, multiplying yields $z_1z_2 \equiv t_1t_2 \pmod{x-y}$, as desired. Notice $(a+c)-(b+d)$ divides $(a+c)^2-(b+d)^2 = [(a^2+c^2)-(b^2+d^2)] + 2(ac-bd)$, since our assumption is that $(a+c)-(b+d)$ divides the first part, it follows that it also divides $2(ac-bd)$, but $(a+c)-(b+d)$ is odd, hence it divides $ac-bd$. We'll prove the claim by induction. For $n=1,2$ it is true. Now let $n \geq 3$ and suppose that $(a+c)-(b+d)$ divides both $(a^{n-1}+c^{n-1})-(b^{n-1}+d^{n-1})$ and $(a^{n-2}+c^{n-2})-(b^{n-2}+d^{n-2})$. We have that $(a+c)(a^{n-1}+c^{n-1}) - (b+d)(b^{n-1}+d^{n-1})$ is divisible by $(a+c)-(b+d)$ (from the Lemma). Expanding gives $(a^n+c^n)-(b^n+d^n)+ac(a^{n-2}+c^{n-2})-bd(b^{n-2}+d^{n-2})$. The sum of the two latter terms is divisible by $(a+c)-(b+d)$ from the Lemma and the inductive hypothesis. Hence $(a+c)-(b+d)$ divides $(a^n+c^n)-(b^n+d^n)$, as claimed.

uglysolutions wrote: Lemma. If $x-y \mid z_1-t_1$ and $x-y \mid z_2-t_2$ then $x-y \mid z_1z_2 - t_1t_2$. Proof. We have that $z_i \equiv t_i \pmod{x-y}$ for $i=1,2$, multiplying yields $z_1z_2 \equiv t_1t_2 \pmod{x-y}$, as desired. Notice $(a+c)-(b+d)$ divides $(a+c)^2-(b+d)^2 = [(a^2+c^2)-(b^2+d^2)] + 2(ac-bd)$, since our assumption is that $(a+c)-(b+d)$ divides the first part, it follows that it also divides $2(ac-bd)$, but $(a+c)-(b+d)$ is odd, hence it divides $ac-bd$. We'll prove the claim by induction. For $n=1,2$ it is true. Now let $n \geq 3$ and suppose that $(a+c)-(b+d)$ divides both $(a^{n-1}+c^{n-1})-(b^{n-1}+d^{n-1})$ and $(a^{n-2}+c^{n-2})-(b^{n-2}+d^{n-2})$. We have that $(a+c)(a^{n-1}+c^{n-1}) - (b+d)(b^{n-1}+d^{n-1})$ is divisible by $(a+c)-(b+d)$ (from the Lemma). Expanding gives $(a^n+c^n)-(b^n+d^n)+ac(a^{n-2}+c^{n-2})-bd(b^{n-2}+d^{n-2})$. The sum of the two latter terms is divisible by $(a+c)-(b+d)$ from the Lemma and the inductive hypothesis. Hence $(a+c)-(b+d)$ divides $(a^n+c^n)-(b^n+d^n)$, as claimed. Beautiful! This was my solution too!

Sea $k=a-b+c-d$. Notemos que $a+c \equiv b+d$ $(mod$ $k)$. Por lo tanto, $(a+c)^{2} \equiv (b+d)^{2}$ $(mod$ $k)$, en consecuencia, $a^{2}+c^{2}+2ac \equiv b^{2}+d^{2}+2bd$ $(mod$ $k)$, usando el dato del problema llegamos a que $2ac \equiv 2bd$ $(mod$ $k)$, pero como $k$ es impar, esto se reduce a que $ac \equiv bd$ $(mod$ $k)$ (a este resultado le llamaremos $(1)$). Luego, Supongamos que existe un entero positivo $n \geq 2$ tal que, para todo $i$, con $1 \leq i \leq n$, se cumpla que $a^{i}+c^{i} \equiv b^{i}+d^{i}$ $(mod$ $k)$ (a esto le llamaremos $(2)$), entonces se cumple que $(a+c)(a^{n}+c^{n}) \equiv (b+d)(b^{n}+d^{n})$ $(mod$ $k)$, en consecuencia, $a^{n+1}+c^{n+1}+ac(a^{n-1}+c^{n-1}) \equiv b^{n+1}+d^{n+1}+bd(b^{n-1}+d^{n-1})$ $(mod$ $k)$, pero es fácil observar usando $(1)$ y $(2)$ que se cumple lo siguiente: $ac(a^{n-1}+c^{n-1}) \equiv bd(b^{n-1}+d^{n-1})$ $(mod$ $k)$. Por lo tanto, $a^{n+1}+c^{n+1} \equiv b^{n+1}+d^{n+1}$ $(mod$ $k)$, que es lo que queremos probar.

Above: literally the same solution as the one from @uglysolutions, just that it's in Spanish.

Let $e=a+b-c-d.$ Notice $$e\mid (a+b-c-d)(a+b+c+d)-(a^2+b^2-c^2-d^2)=2(ac-bd)$$but $e$ is odd so $e\mid ac-bd.$ We proceed by strong induction to show $e\mid a^n-b^n+c^n-d^n.$ Notice $e\mid e$ and $e\mid a^2+c^2-b^2-d^2$ imply the base cases $n=1,2,$ respectively. Assume $n=k-2,k-1$ work. Then, \begin{align*}a^k+c^k+ac(a^{k-2}+c^{k-2})&\equiv (a+c)(a^{k-1}+c^{k-1})\\&\equiv (b+d)(b^{k-1}+d^{k-1})\\&\equiv b^k+c^k+bd(b^{k-2}+d^{k-2}).\end{align*}$\square$

we have: $a-b+c-d=2k+1$ and $2k+1|a^2-b^2+c^2-d^2$ $\Longrightarrow a+c \equiv b+d (mod 2k+1) ...(1)$ $\Longrightarrow a^2+c^2 \equiv b^2+d^2 (mod 2k+1)$ $\Longrightarrow ac \equiv bd (mod 2k+1)$ Let's suppose $a^m+c^m \equiv b^m+d^m (mod 2k+1)$ up to $m=n-1$, let's see if this is true for $m=n$: $\Longrightarrow a^{n-1}+c^{n-1} \equiv b^{n-1}+d^{n-1} (mod 2k+1)...(2)$ of $(1)$ and $(2)$: $\Longrightarrow c^n+a^n+a^{n-1}c+ac^{n-1} \equiv b^n+b^{n-1}d+bd^{n-1}+d^n (mod 2k+1)$ $\Longrightarrow a^n+c^n+ac(a^{n-2}+c^{n-2}) \equiv b^n+d^n+bd(b^{n-2}+d^{n-2})(mod 2k+1)$ $\Longrightarrow a^n+c^n \equiv b^n+d^n (mod 2k+1)$ then the induction is complete thus $a-b+c-d|a^n-b^n+c^n-d^n$