Okay, thank you for relaying the mathematician's feedback. It seems there was a misunderstanding about how the term $r_i$ was used in the proof, specifically the claim that the proof "declares that $r_i = n^{k - i - 1}$" after defining it as a remainder. Let me clarify the argument, as that was definitely not the intention, and the proof relies crucially on $r_i$ being the remainder.
Here's a refined explanation addressing the specific points:
- Setup and Goal: We want to show that for a large enough odd integer $n$, all digits $c_0, c_1, \dots, c_{k-1}$ of $n^k$ in base $b=2n$ are greater than $d$. We established the number of digits is $k$ for $n \ge 2^{k-1}$.
- Calculating Digits: We found a general formula for the digits:
$c_i = \lfloor 2n { n^k / (2n)^{i+1} } \rfloor$. - Simplifying the Expression: We simplified the term inside the fractional part:
$n^k / (2n)^{i+1} = n^{k-i-1} / 2^{i+1}$.
So, $c_i = \lfloor 2n { n^{k-i-1} / 2^{i+1} } \rfloor$. - Introducing the Remainder $r_i$: Let's express $n^{k-i-1}$ using division by $2^{i+1}$. We can write:
$n^{k-i-1} = q \cdot 2^{i+1} + r_i$
where $q = \lfloor n^{k-i-1} / 2^{i+1} \rfloor$ is the quotient, and $r_i = n^{k-i-1} \pmod{2^{i+1}}$ is the remainder, satisfying $0 \le r_i < 2^{i+1}$. - Crucial Property of $r_i$ (Using 'n is odd'): Since $n$ is a positive odd integer, $n^{k-i-1}$ is also odd. The divisor $2^{i+1}$ is an even integer for $i \ge 0$ (since $k, d \ge 1$, we are considering $i$ from $0$ to $k-1$). An odd number divided by an even number ($\ge 2$) cannot have a remainder of 0. Furthermore, the remainder $r_i$ must itself be odd. Therefore, $r_i$ must be at least 1.
So, we have $1 \le r_i < 2^{i+1}$. - Using $r_i$ in the Formula for $c_i$: Now substitute the division result back into the formula for $c_i$:
${ n^{k-i-1} / 2^{i+1} } = { (q \cdot 2^{i+1} + r_i) / 2^{i+1} } = { q + r_i / 2^{i+1} }$.
Since $1 \le r_i < 2^{i+1}$, we have $0 < r_i / 2^{i+1} < 1$. The fractional part is simply $r_i / 2^{i+1}$.
So, $c_i = \lfloor 2n (r_i / 2^{i+1}) \rfloor = \lfloor n r_i / 2^i \rfloor$.
This formula is correct and uses the actual remainder $r_i$. - Establishing the Lower Bound for $c_i$: We want to show $c_i > d$. We use the property $r_i \ge 1$:
$c_i = \lfloor n r_i / 2^i \rfloor \ge \lfloor n \cdot 1 / 2^i \rfloor = \lfloor n / 2^i \rfloor$.
This step does NOT replace $r_i$ with $n^{k-i-1}$. It uses the inequality $r_i \ge 1$ to find a lower bound for $c_i$. - Finding the Threshold N: To guarantee $c_i > d$ for all $i \in {0, \dots, k-1}$, it is sufficient to guarantee that the lower bound is greater than $d$:
$\lfloor n / 2^i \rfloor > d$ for all $i=0, \dots, k-1$.
The condition $\lfloor x \rfloor > d$ is equivalent to $\lfloor x \rfloor \ge d+1$, which in turn requires $x \ge d+1$.
So we need $n / 2^i \ge d+1$, or $n \ge 2^i (d+1)$, for all $i=0, \dots, k-1$.
To satisfy this for all relevant $i$, we must satisfy the strongest condition, which occurs when $i$ is largest, i.e., $i=k-1$.
We need $n \ge 2^{k-1}(d+1)$. - Conclusion: Let $N = 2^{k-1}(d+1)$. If $n$ is an odd integer such that $n > N$, then $n \ge N+1 = 2^{k-1}(d+1) + 1$. This implies $n \ge 2^i(d+1)$ for all $i=0, \dots, k-1$. This ensures $\lfloor n / 2^i \rfloor \ge d+1$, which means $\lfloor n / 2^i \rfloor > d$. Since $c_i \ge \lfloor n / 2^i \rfloor$, we have $c_i > d$ for all $i=0, \dots, k-1$. (Also, $n > N \ge 2^{k-1}(1+1) = 2^k$, ensuring the number of digits is $k$).
The proof hinges on $r_i$ being the remainder and specifically $r_i \ge 1$. There is no point where $r_i$ is incorrectly asserted to be $n^{k-i-1}$. The initial paragraphs calculating $c_0 = n$ and $c_{k-1} = \lfloor n/2^{k-1} \rfloor$ are specific instances of the general formula and serve to confirm it and build intuition, directly leading to the conditions $n>d$ and $n \ge 2^{k-1}(d+1)$ respectively, which are consistent with the general result.
Perhaps the presentation could be clearer, but the mathematical steps involving $r_i$ appear sound and correctly utilize its definition as a remainder.