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Let $$N = 2007^{2013}-1974^{2013}-1946^{2013}+1913^{2013}$$ Then select all the option(s) that are correct:

  1. N is divisible by 61
  2. N is divisible by 2013
  3. N is divisible by 28
  4. All of these

My attempt: I tried to use the property $a^x - b^x = (a-b)(a^{x-1} + ... + b^{x-1})$ for odd x. Note that $2007 - 1946 = 61$ and $1974-1913 = 61$ $$N = 61(2007^{2012} +\ ...\ + 1946^{2012} - (1974^{2012} +\ ...\ + 1913^{2012})) $$ Option 1 is correct. However, the answer key says that $N$ is also divisible by $2013$. How do I prove this? Little fermat won't work because $2013$ is not prime.

Michael Rozenberg
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Aniruddha Deb
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    You can use Euler's theorem as a generalization of Fermat's little theorem : If $a$ and $n$ are coprime, then $$a^{\varphi(n)}\equiv 1\mod n$$ holds. – Peter Sep 18 '20 at 10:05
  • But if you have already established that the given number is divisible by $61$ , you only need to do that also for $11$ and $3$ considering $2013=3\cdot 11\cdot 61$ and the chinese remainder theorem. – Peter Sep 18 '20 at 10:08
  • I added an answer that shows how it is a special case of a general result that makes the divisibilities obvious. – Bill Dubuque Sep 18 '20 at 15:38

2 Answers2

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We have $$2013=61\cdot33.$$ Since $$2007-1946=1974-1913=61$$ and $$N=2007^{2013}-1946^{2013}-\left(1974^{2013}-1913^{2013}\right),$$ we see that $N$ is divisible by $61$.

Thus, it's enough to prove that $N$ is divisible by $33.$

Now, write $$N=2007^{2013}-1974^{2013}-\left(1946^{2013}-1913^{2013}\right).$$ Can you end it now?

Also, since $$N=2007^{2013}+1913^{2013}-\left(1946^{2013}+1974^{2013}\right),$$ we see that $N$ is divisible by $28$.

Michael Rozenberg
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Corect is "All" by $\,m,n,d =\!\!\!\!\overbrace{ 61,33}^{\color{darkorange}{61(33)=2013}}\!\!\!\!,1913\,$ below, $ $ & $\bmod \color{darkorange}{28}\!:\ \color{#0a0}{m\!+\!n\!+\!2d}\equiv 5\!+\!5\!+\!2\cdot 9\equiv 0$

$$\begin{align} f\, =\ &\ \ \ \ \ \ \ \ \ \ 2007^k -\ \ \ \ \ 1974^k -\ \ \ 1946^k +1913^k\\[.1em] =\ &(\color{#c00}m\!+\!n\!+\!d)^k-(\color{#c00}m\!+\!d)^k-(n\!+\!d)^k+d^k\\[.2em] {\large \Rightarrow}\ \ f\,\equiv\ &\ \ \ \ \ \ \ (n\!+\!d)^k \ \ \ \ -\ \ \ \ \ d^k\: - (n\!+\!d)^k + d^k\equiv 0\pmod{\!\color{#c00}m\!=\!\color{darkorange}{61}}\\[.1em] \&\ \ \ f\,\equiv\ &\ \ \ \ \ \ (m\!+\!d)^k - (m\!+\!d)^k \ \ \ \ - \ \ \ d^k\ +\ d^k\equiv 0\pmod{n\!=\!\color{darkorange}{33}}\\[.1em] \&\ \ \ f\,\equiv\ &\ \ \ \ \ \ \ \ \ \ (-d)^k\!\! +\! (-n\!-\!d)^k\! + (n\!+\!d)^k +\, d^k\equiv 0\pmod{\underbrace{\color{#0a0}{m\!+\!n\!+\!2d}}_{\textstyle \color{darkorange}{28}k\!\!\!}},\ {\rm by}\ \ k\ \rm odd \end{align}\qquad$$

Remark $ $ Thus for any integers $\,m,n,d\,$ the above $\,f(m,n,d)\,$is divisible by $m,n$, and also by $\,m\!+\!n\!+\!2d$ if $k$ is odd. See this answer for more on the innate symmetry at the heart of the matter.

Bill Dubuque
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  • I upvoted, very interesting. With respect to my answer : (1) good catch (2) Your answer does focus on a general approach, making my answer obselete. Therefore, I am going to delete my answer. – user2661923 Sep 18 '20 at 16:20
  • @user2661923 Thanks. Once the underlying symmetry is brought to the fore (see the linked post) it becomes obvious how to generalize such problems (and, alas, as is often the case, the "magic" disappears once one learns the "trick" and it appears then a bit trivial). But at least we can still appreciate the power that symmetry brings to the table - which is beautiful in itself (and more ubiquitous). As I often advise my students: always exploit any innate symmetry! – Bill Dubuque Sep 18 '20 at 18:53