Proving modular statements .

Question

Prove:

a) $$\left( n+1\right) ^{n}-1\equiv 0 \pmod{n^2}$$

b) $$ \left( n+2\right) ^{n+2}-2^{n+2}\left( n+1\right) ^{n+1}\equiv 0\pmod{n^2}$$

@DamilaWhiShadow Add that to your question, else your answer may be closed or put on hold as off-topic like this one — Jaideep Khare, Apr 28 '17 at 22:57
But this is a humble and remote attempt ,it's not worth being mentioned. — Am ine, Apr 28 '17 at 23:03
@DamilaWhiShadow You mean that $(n+1)^n \equiv 1 ; \mathrm{mod}; n$? — Sentinel135, Apr 28 '17 at 23:33
No it's $$\left( n+1\right) ^{n}\equiv 1\left[ n^{2}\right]$$ — Am ine, Apr 28 '17 at 23:37
Well if you have that. then what's the problem? Or rather what's $1$ in $\mathrm {mod}; n^2$? — Sentinel135, Apr 28 '17 at 23:39
B.T.W. $a ; \mathrm {mod}; b$ isn't normally written as $a[b]$, but rather written as $[a]_b$. This form is more understandable. — Sentinel135, Apr 28 '17 at 23:41

score 2 · Accepted Answer · answered Apr 28 '17 at 23:34

I would suggest you to try to use the binomial theorem expansion which can be proved by induction, for more details about that see https://en.wikipedia.org/wiki/Binomial_theorem. At least on part a) you should get: $$ \sum_0^n \frac{n!}{(n-k)!(k!)}n^k-1 $$ which should expand to something like: $$n^n+nn^{(n-1)}+...+n^2+1-1=0\ mod \ n^2$$ Hopefully using the same Binomial expansion theorem you should get something useful for the second part.

Proving modular statements .

1 Answers1