I am going through the book of proof by Richard Hammack and came across this exercise:
Prove that if $n=2^{k}-1$ for $k\in\mathbb{N}$, then every entry in Row $n$ of Pascal's Triangle is odd.
My proof is the following:
Proof.(Contrapositive) Suppose it is not the case that every entry in Row $n$ of Pascal's Triangle is odd. Thus, there is a entry in Row $n$ of Pascal's Triangle that is even.
Lemma: If there is a entry in Row $n$ of Pascal's Triangle that is even, there are two consecutive entries such that one is even and the other one is odd.
Proof of Lemma:(Contrapositive) Suppose it is not the case that there are two consecutive entries such that one is even and the other one is odd. Thus, the entries in Row $n$ of Pascal's Triangle are either all even or all odd. All the entries in Row $n$ of Pascal's Triangle are obviously not even, since for all $n\in\mathbb{N}$, we have that $\binom{n}{0}$ and $\binom{n}{n}$ are equal to $1$ which is odd. Thus, it must be true that all the entries in Row $n$ of Pascal's Triangle are odd. Therefore, it is not the case that there is a entry in Row $n$ of Pascal's Triangle that is even. $\square$
Because of our Lemma, there are two consecutive entries such that one is even and the other one is odd. Let $m\in\mathbb{N},0\leq m\leq n$. Without loss of generality, let $\binom{n}{m}$ and $\binom{n}{m+1}$ be consecutive entries of Row $n$ of Pascal's Triangle, such that $\binom{n}{m}$ is even and $\binom{n}{m+1}$ is odd.
It is a known relation that $\binom{n+1}{m+1}=\binom{n}{m}+\binom{n}{m+1}$. Since $\binom{n}{m}$ is even and $\binom{n}{m+1}$ is odd, it follows that $\binom{n+1}{m+1}$ is odd.
Notice that \begin{align} \binom{n+1}{m+1}&=\frac{(n+1)!}{(m+1)!(n-m)!}\nonumber\\ &=\frac{(n+1)n!}{(m+1)m!(n-m)!}\nonumber\\ &=\frac{(n+1)}{(m+1)}\binom{n}{m}.\nonumber \end{align}
This means $\binom{n+1}{m+1}=\frac{(n+1)}{(m+1)}\binom{n}{m}$. But $\binom{n+1}{m+1}$ is odd and $\binom{n}{m}$ is even. This can only be true if $m+1$ is even and has exactly as many factors of $2$ as $(n+1)\binom{n}{m}$. Suppose $\binom{n}{m}=2^{a}b$ for $a,b\in\mathbb{Z}$ such that $b$ is odd. Thus, $m+1=2^{c}d$, for $c,d\in\mathbb{Z}$ such that $d$ is odd.
Suppose $n+1=2^k$. It follows that $c=k+a$ and $m+1=2^{k+a}d>2^k=n+1$. Thus, $m>n$ which is a contradiction.
Thus, $n\neq2^{k}-1$ and $k\in\mathbb{N}$. Therefore, it is not the case that $n=2^{k}-1$ for $k\in\mathbb{N}$. $\blacksquare$
Is this correct? I was able to find a similar question posted here
If $n = 2^k - 1$ for $k \in \mathbb{N}$, then every entry in row $n$ of pascal's triangle is odd.
but the proof and the comments went on a different aproach (using Lucas Theorem, or some math I wasn't introduced to), but since the exercise was on a book that requires little to no previous knowlegde, and all that was introduced so far was direct and contrapositive proof, I can't imagine the author wanted me to use some advanced theorem or definition.
