I have the following problem: given a prime number $p_k$ and the prime immediately following $p_{k+1}$, is it possible to find a prime number $q$, with $q\ne p_k$ and $q\ne p_{k+1}$ such that the following equation holds? $\frac{1}{p_k}+\frac{1}{p_{k+1}}=\frac{1}{q}$ Thanks in advance.
Equation involving reciprocal of prime numbers
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1The same idea can be used to show that if $p$ and $q$ are **any** primes, then the only way that we can have $\frac{1}{p}+\frac{1}{q}=\frac{1}{n}$, where $n$ is an **integer**, is $p=q=2$ or $p=2$, $q=3$, or $p=3$, $q=2$. – 2012-10-11
3 Answers
You can rewrite the equation as \begin{equation} q(p_{k} + p_{k + 1}) = p_{k} p_{k + 1}. \end{equation} Hence, $ q $ must divide either $ p_{k} $ or $ p_{k + 1} $, since $ q $ is a prime number. It follows that either $ q = p_{k} $ or $ q = p_{k + 1} $, so either $ \dfrac{1}{p_{k + 1}} = 0 $ or $ \dfrac{1}{p_{k}} = 0 $ respectively. Contradiction. The equation, therefore, cannot be solved.
No. If you multiply out the fractions, you get $p_{k+1}q + p_kq = p_kp_{k+1}$ Then you can see that the prime $p_k$ divides two of the terms, and hence would divide the other, and you get an immediate contradiction.
More can be said. Suppose $p,q$ and $r$ are primes. Then $\frac1p + \frac1q = \frac1r$ cannot hold.
Suppose it does. Then $r = \frac{pq}{p+q}.$ But the only divisors of $pq$ are $1,p,q$ and $pq$. Clearly, $p+q$ cannot equal the first three of these. So the only remaining possibility is $p+q=pq$. But in this case $r$ must be $1$, which is not prime.