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\title{Assignment 7}%replace X with the appropriate number
\author{Jennifer Wu\\ %replace with your name
Math 135 - Prof Haykazyan} %if necessary, replace with your course title

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\begin{question}{1}
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Note that because p is prime (so $p > 0$) and k is defined as $0 \leq k < p$, by the Exercise on Page 74, $\binom{p}{k}$ is a non-negative integer. Also note that $\binom{p}{k} = \frac{p!}{(p-k)!k!}$, and as $p$ is prime, it can't be "canceled out" by any factors in the denominator $(p-k)!k!$. So $\frac{p!}{p(p-k)!k!} = \frac{(p-1)!}{(p-k)!k!}$ must be an integer. Thus, we know that $p \mid \frac{p!}{(p-k)!k!}$ and so $p \mid \binom{p}{k}$. From this, since $a \in \mathds{Z}$, $a^k \in \mathds{Z}$, and so we also know $p \mid \binom{p}{k}a^k$.
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And so, because every element in $\sum_{k=1}^{p-1} \binom{p}{k}a^k$ is divisible by $p$, $\sum_{k=1}^{p-1} \binom{p}{k}a^k \equiv 0 \text{ (mod p)}$. We also know from BT1 that \[\sum_{k=1}^{p-1} \binom{p}{k}a^k = \sum_{k=0}^{p} \binom{p}{k}a^k - 1 - a^p = (1+a)^p - 1 - a^p\] And because we know that $\sum_{k=1}^{p-1} \binom{p}{k}a^k \equiv 0 \text{ (mod p)}$, we also know that $(1+a)^p - 1 - a^p \equiv 0 \text{ (mod p)}$. By CAM, we add $1 + a^p \equiv 1 + a^p$ (mod p) to both side to get \[(1+a)^p \equiv 1 + a^p \text{ (mod p)}\] and we are done.
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