0

Show the number $N = 760^{1998} - 20^{1998} + 1910^{1998} - 652^{1998}$ is divisible by 1998. I tried to factor in the numbers and use Euler's theorem as I am applying?

Bill Dubuque
  • 272,048
trombho
  • 1,591
  • Note: Google translates this as "

    How to solve it? Show the number $N = 760 ^ {1998} - 20 ^ {1998} + 1910 ^ {998} - 652 ^ {19998}$ is divisible by $1998$

    I tried to factor in the numbers and use Euler's theorem as I am applying?"

     – lulu Sep 29 '19 at 16:02
  • Other Note: I question some of these exponents. As it stands the claim does not appear to be true. WA gets the result $678$. – lulu Sep 29 '19 at 16:04
  • Please see https://math.meta.stackexchange.com/questions/5020/ – Angina Seng Sep 29 '19 at 16:06
  • I've already corrected! Forgiveness for what happenedI've already corrected! Forgiveness for what happened – trombho Sep 29 '19 at 16:15
  • I added an answer exploiting symmetry. You can find many similar examples in prior answers. – Bill Dubuque Sep 29 '19 at 21:27

2 Answers2

2

The problem enjoys innate $\rm\color{#c00}{symmetry}$ that greatly simplifies matters once brought to the fore.

$\quad\phantom{\Rightarrow}\ \ \{ 760,\ \ \ \ 1910\}\ \ \equiv\, \{20,\ \ \ 652\}\ \ \ \ {\rm mod}\,\ 27\ \&\ 37,\ $ by $\ \ \begin{align}&760,\,1910\,\equiv\, 652,\, 20\,\bmod{27}\\ &760,\,1910\,\equiv\, 20,\, 652\,\bmod{37}\end{align}$

$\quad\Rightarrow\ \{760^n,\ \ 1910^n\} \equiv \{20^n,\ \,652^n\}\,\ {\rm mod}\,\ 27 \ \&\ 37,\ $ by the Congruence Power Rule

$\quad\Rightarrow\ \ \ 760^n\!+\! 1910^n\ \ \equiv \ \ 20^n\!+652^n\ \ \,{\rm mod}\,\ 27\ \&\ 37,\ $ so also $\,{\rm mod}\ 999 = {\rm lcm}(27,37) = 27(37)$

since addition $\,f(x,y)\, =\, x + y\ $ is $\rm\color{#c00}{symmetric}$ $\,f(x,y)= f(y,x),\, $ so its value depends only upon the (multi-)set $\,\{x,\ y\}.\, $ Parity $\,\Rightarrow\,$ congruence is true mod $2,\,$ so also mod $\,2\cdot 999 = 1998.\,\ $ QED

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

$$\begin{align} \{A, B\}\, &\equiv\, \{a,b\}\ \pmod{\ m\ \ \&\ \ n}\\[.5em] \Rightarrow\ f(A,B)&\equiv f(a,b)\, \pmod{{\rm lcm}(m,n)}\end{align}\qquad\qquad$$

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

You can find many similar symmetric problems in prior answers.

Bill Dubuque
  • 272,048
1

Note that $1998=2\times 27\times 37$ . $N = 760^{1998} - 20^{1998} + 1910^{1998} - 652^{1998}$ Since all the constituents of N are even, It is enough to prove that $999\mid N$, which is equivalent to proving that $27\mid N$ and $37\mid N$ as $(27,37)=1$. Note that, $$760\equiv 20 \pmod {37}$$ and $$1910\equiv 652 \pmod {37}$$. Hence $760^{1998}\equiv 20^{1998}\pmod {37}$ and $1910^{1998}\equiv 652^{1998} \pmod {37}$. Also note that $\varphi(27)=18$ and $18\mid 1998$ and $(760,27)=(20,27)=(1910,27)=(652,27)=1$. Now use Euler's theorem to complete the proof. I think you can take it from here.