Problem using binominal theorem?

Question

I tried to solve this problem but I couldn't so i'm looking for an help.

Is there any two digit natural number $n$ which fits with following statement? $$n \mid (4^n - 3^n - 1)$$

The hint is using the binomial theorem.

Answer 1

Note that by the binomial theorem $$4^n=(3+1)^n=3^n+\sum_{k=1}^{n-1}\binom{n}{k}\cdot 3^k\cdot 1^{n-k}+1^n.$$ Hence
$$4^n-3^n-1=\sum_{k=1}^{n-1}\binom{n}{k}\cdot 3^k.$$ Show that for $n=11$ (any prime will work) then the binomial coefficients $\binom{11}{k}$ for $k=1,\dots,10$ are all multiple of $11$.

Answer 2

$4^n-3^n-1=(1+3)^n-3^n-1 = \sum\limits_{i=0}^n3^i\binom{n}{i} -n^n-1= \sum\limits_{i=1}^{n-1}3^{i}\binom{n}{i}$.

If we plug in a prime number for $n$ then $\binom{n}{i}$ is always a multiple of $n$ for $1\leq i \leq n-1$

Answer 3

Try $n=15$:

$$(3+1)^{n}-3^{n}-1 = 3^n+{n\choose 1}3^{n-1}+...+{n\choose n-1}3+1-3^n-1$$ $$= {n\choose 1}3^{n-1}+...+{n\choose n-1}3 $$ $$= n\cdot 3^{n-1}+...+n\cdot 3 =15k$$

Answer 4

Any prime value works by fermats little theorem:

$4^p-3^p-1\equiv 4-3-1\bmod p$

Answer 5

Any (two digit, since you specify it) prime $p$ satisfies this condition, since the condition can be re-written as $$4^n=(3+1)^n\equiv 3^n+1^n\mod n$$ and lil' Fermat just stipulates that if $p$ is prime, $x^p\equiv x\mod p$ for any $x$, so $$(3+1)^p\equiv 3+1\equiv 3^p+1^p\mod p.$$