$((1+(pq+1)+(pq+1)^2 ....(pq+1)^{p-1})$ is divide by $p^2$($p$ is prime and $p$ and $q$ is coprime number)?
Note that $((1+(pq+1)+(pq+1)^2 ....(pq+1)^{p-1})$ is divide by $p$, ($p$ and $q$ are coprime and $p$ is prime) .
Prove that $((1+(pq+1)+(pq+1)^2 ............(pq+1)^{p-1})$ is divide by $p^2$($p$ is prime and $p$ and $q$ is coprime number).
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number-theory
elementary-number-theory
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0Expand with this good old binomial law. Do you see something ? – 2017-01-04
1 Answers
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It is false. For $p=3$ and $q=2$ we obtain $pq+1=7$. Then the sum is $1+7+7^2=57$, which is a multiple of $p$ but not of $p^2$.
In fact this is what always happens:
$$\sum_{k=0}^{p-1}(1+pq)^{k}=\frac{(pq+1)^p-1}{pq+1-1}=\frac{p^3q^2A+p^2q}{pq}=p^2qA+p$$ where $A$ is some integer.