$ 1987 \mid \left( n^n + (n+1)^n \right) $

Question

Problem from the 1987 Leningrad Math Olympiad:

Is there a positive integer $n$ such that $ n^n + \left( n + 1 \right)^n $ is divisible by $ 1987 $?

The provided solution:

The answer is yes. Take $ n = 993 $ and we get $ \left( n + n + 1 \right) \mid \left( n^n + (n+1)^n \right) $, so $ 2n + 1 = 1997 $, which gives $ n = 993 $.

My question is: really? How is it obvious that $ \left( n + n + 1 \right) \mid \left( n^n + (n+1)^n \right) $? Is that a true statement for all $n$? If so, what is the proof? Thanks!

Notice $k+ 1 = -k \pmod {2k+1}$. If $s$ is an odd number, you get $k^s + (k+1)^s = k^s + (-k)^s = 0 \pmod {2k+1}$ — achille hui, Aug 11 '14 at 16:57
possible duplicate of Proof of $a^n+b^n$ divisible by a+b when n is odd — Jyrki Lahtonen, Aug 11 '14 at 16:57
Well. This question (at least without the solution) is way more interesting than the one I proffered this to be a duplicate of. Having second thoughts, but... — Jyrki Lahtonen, Aug 11 '14 at 17:11

score 7 · Accepted Answer · answered Aug 11 '14 at 16:53

7

The key is the following known observation: If $n=2k+1$ then $$a^{n}+b^n=a^{2k+1}+b^{2k+1}=(a+b)( \mbox{junk} )$$ Thus, for all $n$ odd, $a+b$ divides $a^n+b^n$.

answered Aug 11 '14 at 16:53

N. S.

132,525

1

Ah, thanks! Why did I not notice this? I will accept in 11 minutes. – Ahaan S. Rungta Aug 11 '14 at 16:54

score 2 · Answer 2 · answered Aug 11 '14 at 16:53

2

$a+b\mid a^k+b^k$ when $k$ is odd.

answered Aug 11 '14 at 16:53

Thomas Andrews

177,126

My question was, what is the proof for this? – Ahaan S. Rungta Aug 11 '14 at 16:53
1

Well, no, your question was, why is $n+(n+1)\mid n^n + (n+1)^n$. I reduced it to a theorem I assumed you'd know :) – Thomas Andrews Aug 11 '14 at 16:54
Okay, fair enough. +1. :) – Ahaan S. Rungta Aug 11 '14 at 16:55
@AhaanRungta $a^k-b^k=(a-b)(a^{k-1}+a^{k-2}b+\cdots+b^{k-1})$ so $a-b\mid a^k-b^k$. If $k$ is odd then $a+b=a-(-b)\mid a^k-(-b)^k=a^k+b^k$ . – drhab Aug 11 '14 at 17:30

score 2 · Answer 3 · edited Apr 13 '17 at 12:21

${\bf Hint}\ \ n-m\mid n^k-m^k.\ $ So $\,\ m=-(n\!+\!1),\,\ k\,$ odd
${\rm yields}\ \ 2n\!+\!1\mid n^k\!+(n\!+\!1)^k.\,\ $ Yours is case $\,k=n.$

Remark $\ $ The above divisibility can be viewed as a special case of the Factor Theorem $\, x-y\,\ \mid\,\ f(x)-f(y)\ $ for any polynomial $\,f\,$ or, by the Polynomial Congruence Theorem $\,x\equiv y\Rightarrow f(x)\equiv f(y),\,$ where above $\,f(x) = x^k.$

$ 1987 \mid \left( n^n + (n+1)^n \right) $

3 Answers3