# More on the Erdős–Straus conjecture

As a reminder, the Erdős–Straus conjecture is: $\forall n \in \mathbb{N}_1:\frac{4}{n+3}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$.

Let ${n+3} = k$, we can see if $k$ is even, then $k=2j$ and $\frac{4}{k} = \frac{4}{2j} = \frac{2}{j} =\frac{1}{2j} + \frac{1}{2j} + \frac{1}{j}$ so the conjecture holds for all even $k$. In fact this holds for $k=2, j=1$ also.

For odds it’s not so simple, though an expansion does exist for the odd numbers of the form $k=4j+3$