-2

How to prove $(n+1)^{m+1}-(-n)^{m+1}$ is divisible by $2n+1$, with $n,m$ positive integers?

I have tried by induction (on $n$) and with the binomial theorem, something like this (have also assumed $m$ odd here):

$((n+1)+1)^{m+1}-(-(n+1))^{m+1}\\=(m+1)n^m+(m+1)(mn^{m-1}+...+mn+1)\\+\frac{(m+1)m}{2!}n^{m-1}+\frac{(m+1)m}{2!}n^{m-1}((m-1)n^{m-2}+...+(m-1)n+1)\\+...\\+\frac{(m+1)m}{2!}n^{2}+\frac{(m+1)m}{2!}(n+1)\\+(m+1)n+(m+1)\\+1$

The terms on the left column are $k(2n+1)$ for some $k$ integer by induction hypothesis, and I would need to somehow rewrite the rest of the terms. Do I need to do induction on $m$ here recursively? Does this direction for the proof make sense at all?

JDd
  • 9
  • You can use induction on both $n$ and $m$. First, prove the base cases for $n=1$ and $m=1$. Then, assume the statement is true for $n$ and $m$, and prove it for $n+1$ and $m$. Finally, assume the statement is true for $n+1$ and $m$, and prove it for $n+1$ and $m+1$. This way, you can use the induction hypothesis to rewrite the terms and show that the expression is divisible by $2n+1$. – rumathe Mar 20 '23 at 17:27

3 Answers3

2

The identity $$x^{m+1} - y^{m+1} = (x-y)\left( \sum_{i=0}^{m}x^iy^{m-i} \right)$$

is easily proved by induction on $m$.

Now, simply substitute $x=n+1$ and $y=-n$ and you get $$(n+1)^{m+1} - (-n)^{m+1} = (2n+1) \cdot (\mathrm{integer}).$$

KCd
  • 46,062
Crostul
  • 36,738
  • 4
  • 36
  • 72
2

Make calculations in modular arithmetic: work modulo $2n+1$.

Check that $n+1 \equiv -n \bmod 2n+1$. Congruent numbers for a modulus are still congruent after raising them to the same power, so $(n+1)^k \equiv (-n)^k \bmod 2n+1$ for all $k \geq 0$. Thus $(n+1)^k - (-n)^k \equiv 0 \bmod 2n+1$, which says $2n+1$ is a factor of $(n+1)^k - (-n)^k$.

KCd
  • 46,062
  • 1
    Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Mar 20 '23 at 19:17
  • Thank you all for your answers. I had the intuition there had to be a better way to prove it, I think this proof is superior to the induction one, don´t you think? – JDd Mar 21 '23 at 12:56
0

Induction on $m$ is the way to go: \begin{eqnarray} (n+1)^{m+1}-(-n)^{m+1} &=& (n+1)^{m}(n+1) + n(n+1)^m - n(n+1)^m+n(-n)^m\\ &=& (2n+1)(n+1)^m - n[(n+1)^m-n^m]. \end{eqnarray} So if $2n+1$ divides $(n+1)^m-n^m$, it also divides $(n+1)^{m+1}-n^{m+1}$. Since $2n+1$ also divides $(n+1)^0-(-n)^0 = 0$, the statement is true for all $m$.

eyeballfrog
  • 22,485