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- Thread starter Poirot
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- Feb 13, 2012

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If $p_{1} \ne 2$ then the term $(p_{1}-1)\ ...\ (p_{k}-1)$ is even and the term $p_{1}\ ...\ p_{k}$ is odd ... that's impossible so that it must be $p_{1}=2$. In this case the term $(p_{1}-1)\ ...\ (p_{k}-1)$ contains as factor $2^{k-1}$ and the term $p_{1}\ ...\ p_{k}$ contains as factor 2, so that it must be k=2. In this case it is...I can't see why the following is true:

Let $p_{1}<p_{2}<......<p_{k}$ be primes such that

$7(p_{1}-1)......(p_{k}-1)=3p_{1}....p_{k}$.

Since 7 divides the LHS, $p_{k}>or =7$

$\displaystyle 7\ (p_{2}-1)= 3\ 2\ p_{2} \implies p_{2}=7$ (2)

Kind regards

$\chi$ $\sigma$