Show $\,1897\mid 2903^n - 803^n - 464^n + 261^n\,$ by induction Show that: $2903^n - 803^n - 464^n + 261^n$ is divisible by $1897$ for all integers $n\geq1$ using induction.
 A: (I missed out the "by induction" in the question when I wrote this answer, but I will leave it here anyway. Do feel free to have it removed.)
I remember this as a classic Olympiad problem for factorization.
We can rewrite the given expression as:
$$(2903^n - 803^n) - (464^n - 261^n)$$
Then, factorizing (using the difference of powers factorization) we get:
$$(2903 - 803)(\dots) - (464 - 261)(\dots)$$
$$= 2100\cdot(\dots) - 203(\dots)$$
which is divisible by $7$. (I've omitted the expression in the bracket, but the important thing is that they are integers)
Next, we can also rewrite the given expression as:
$$(2903^n - 464^n) - (803^n - 261^n)$$
Factorizing, we get:
$$(2903 - 464)(\dots) - (803 - 261)(\dots)$$
$$=2439\cdot(\dots) - 542\cdot(\dots)$$
which is divisible by $271$ (because $271$ divides both $2439$ and $542$).
Since both $7$ and $271$ divides the given expression, it follows that $7\cdot271 = 1897$ must also divide it.
A: Hint $\ $ Exploit innate $\rm\color{#c00}{symmetry}$! $ $  Consider a simpler analogous example
$\qquad\phantom{\Rightarrow}\ \  \{ 52,\ \ \ \ 23\}\ \  \equiv\, \{2,\ \ \ 3\}\ \ \ \ {\rm mod}\,\ 10,7,\ \ $   
$\qquad\Rightarrow\ \{52^n,\ \ 23^n\} \equiv \{2^n,\ \,3^n\}\,\ {\rm mod}\,\ 10,7,\ \ $ by the Congruence Power Rule 
$\qquad\Rightarrow\ \ \ 52^n\!+\! 23^n\ \ \equiv \ \ 2^n\!+3^n\ \ \,{\rm mod}\,\ 10,7,\ \ $ so also $\,{\rm mod}\ 70 = {\rm lcm}(10,7)$ 
since addition $\,f(x,y)\, =\, x + y\ $ is $\rm\color{#c00}{symmetric}$  $\,f(x,y)= f(y,x),\ $  therefore its value depends only upon the (multi-)set $\,\{x,\,y\}.\ $  Your problem is completely analogous since
$\qquad\!\phantom{\Rightarrow}  \{2903,\, 261\}\ \equiv\ \{803,\, 464\}\,\ {\rm mod}\,\ 271,7,\ \,  $ where $\,\ 271\cdot 7 = 1897$
Generally $ $ if a polynomial $\,f\in\Bbb Z[x,y]\,$ is $\rm\color{#c00}{symmetric}$ then
$\qquad\qquad\quad \{A, B\}\, \equiv\, \{a,b\}\,\ {\rm mod}\,\ m,\, n\,\ \Rightarrow\,\  f(A,B)\equiv f(a,b)\, \pmod{\!{\rm lcm}(m,n)}\qquad\quad$
a generalization of the constant-case optimization  of CRT = Chinese Remainder,  combined with a generalization of the Polynomial Congruence Rule to (symmetric) bivariate polynomials.
