# Equation involving reciprocal of prime numbers

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.

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There are many approaches. For example, if our two primes $p_k$ and $p_{k+1}$ are odd, bringing the left side to a common denominator makes the numerator even. –  André Nicolas Oct 11 '12 at 16:43
The 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 $p=3$, $q=2$. –  André Nicolas Oct 11 '12 at 17:38

You can rewrite the equation as $$q(p_{k} + p_{k + 1}) = p_{k} p_{k + 1}.$$ 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.

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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.

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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.

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