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Prove that :- $A_n = 2903^n - 803^n - 464^n + 261^n$ is divisible by $1897$ for $n \in \mathbb{N}$

I tried induction but got no where :(.

$p(n)$ : $A_n = 2903^n - 803^n - 464^n + 261^n \mod 1897 = 0 $

for $p(1):$ the statement is true as $A_1 = 1897 \mod 1897 = 0$

let $p(k)$ : $A_k = 2903^k - 803^k - 464^k + 261^k \mod 1897 = 0 $

$$A_{k+1} = 2903^{k+1} - 803^{k+1} - 464^{k+1} + 261^{k+1}$$

$$\begin{align} A_{k+1}\mod 1897 & = 2903^{k+1} - 803^{k+1} - 464^{k+1} + 261^{k+1} \mod 1897 \\ &= 2903^{k+1} + - 803^{k+1} - 464^{k+1} + 261^{k+1} \mod 1897 \\ &= (1006 \times (2903^{k} \mod 1897) \\ & - 803\times (803^{k} \mod 1897) - 464\times (464^{k} \mod 1897) \\ & + 261 \times (261^{k} \mod 1897)) \mod 1897 \end{align}$$

Now i don't know what to do or even if what i did is correct or not. Hints are fine with me. thanks.

2 Answers2

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$1897=7\times 271$, so prove it $\bmod 7$ and $\bmod 271$.

first we do $\bmod 7$:

$2903^n-803^n-464^n+261^n\equiv 5^n-5^n-2^n+2^ n\equiv 0 \bmod 7$

Now we do it $\bmod 271$:

$2903^n-803^n-464^n+261^n\equiv 193^n- 261^n-193^n+261^n\equiv 0 \bmod 271$

Asinomás
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  • Thanks. Just one question, are my induction steps correct ? –  Aug 20 '16 at 20:31
  • it looks fine, except for that $1006$ in there, but I don't think that's gonna help you arrive at a proof. – Asinomás Aug 20 '16 at 20:33
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    @CarryonSmiling If you look closely at the calculations you will see some underlying symmetry. I show how that governs the cancellations in my answer. – Bill Dubuque Aug 20 '16 at 22:06
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The problem has some innate $\rm\color{#c00}{symmetry}$ that greatly simplifies matters once brought to the fore. As a hint, let's examine a simpler analogous example

$\qquad\phantom{\Rightarrow}\ \ \{ 52,\ \ \ \ 23\}\ \ \equiv\, \{2,\ \ \ 3\}\ \ \ \ {\rm mod}\,\ 7\ \&\ 10,\ $ since $\ \ \begin{align}&52,\,23\,\equiv\, 3,\, 2\,\bmod{7}\\ &52,\,23\,\equiv\, 2,\, 3\,\bmod{10}\end{align}$

$\qquad\Rightarrow\ \{52^n,\ \ 23^n\} \equiv \{2^n,\ \,3^n\}\,\ {\rm mod}\,\ 7 \ \&\ 10,\ $ by the Congruence Power Rule

$\qquad\Rightarrow\ \ \ 52^n\!+\! 23^n\ \ \equiv \ \ 2^n\!+3^n\ \ \,{\rm mod}\,\ 7\ \&\ 10,\ $ so also $\,{\rm mod}\ 70 = {\rm lcm}(7,10)$

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}\,\ 7\ \&\ 271\ \, $ where $\,\ 7\cdot 271 = 1897$

Generally $ $ if a polynomial $\,f\in\Bbb Z[x,y]\,$ is $\rm\color{#c00}{symmetric}$ then

$\qquad\qquad\quad \{A, B\}\, \equiv\, \{a,b\}\,\ {\rm mod}\,\ m\ \&\ n\ \Rightarrow\ f(A,B)\equiv f(a,b)\, \bmod{\,{\rm lcm}(m,n)}$

a generalization of the constant-case optimization of CRT = Chinese Remainder, combined with a generalization of the Polynomial Congruence Rule to (symmetric) bivariate polynomials.

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Bill Dubuque
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