# Solutions to Diophantine equation $\frac{1}{n}+\frac{1}{p}=\frac{1}{N}$

For each prime $$p$$ there seems to be a uniqe solution $$n=(p-1)p$$ to the Diophantic equation $$\frac{1}{n}+\frac{1}{p}=\frac{1}{N}$$.

Is that right and if so, how to prove the unicity?

In spite of my 10k+ reputation I have no idea how to try to prove the unicity and "reciprocal diophantine equation" doesn't seems to be efficient search keys.

My only context to this is that I found the question while investigating when $$a+b|ab$$.

I may not be a typical user of MSE. I wasn't talented at or interested in number theory as young and now some 50 years later I have become interested in computational mathematics. Now I've studied some elementaries and make some slow progress, but I'll never be competitive with the typical MSE user.

So please, have patience with me, when I am using your skills!

We have: $$p(n-N)=Nn$$, so $$p|N$$ or $$p|n$$.
If $$p|N$$, let $$N=pk$$ and we get $$n-pk=kn$$, which is a contradiction as LHS$$<$$RHS.
So let $$n=pk$$, which gives $$pk=N(k+1)$$, hence $$k|N$$ and $$(k+1)|p$$, forcing $$k=p-1$$ and $$N=k=p-1$$.