Erdős–Straus conjecture

Can every fraction $4/n$, for a whole number $n$ of at least 2, be split into exactly three fractions whose numerators are 1? For example, $4/5=1/2+1/4+1/20$. The denominators must be positive whole numbers, but they may repeat: $4/2=1+1/2+1/2$. The conjecture says some three-term split works for every eligible $n$.[1]

Can a fixed three-term bound represent $4/n$ using unit fractions for every positive integer denominator $n\ge2$? The required equation is

$$\frac4n=\frac1a+\frac1b+\frac1c,\qquad a,b,c\in\mathbb{Z}_{>0}.$$

The values of $a,b,c$ may depend on $n$ and need not be distinct. For $n=5$, the denominators $2,4,20$ work. Clearing denominators gives $4abc=n(ab+ac+bc)$, a polynomial equation whose positive integer solutions are required separately for each $n$. A four-term sum of $1/n$ is immediate and does not settle this stricter bound.[1]

Many cases work. No proof covers every n.[1]

The checked 2026 prime-denominator criterion concerns a particular representation type. It does not settle every denominator; finite checks cannot supply the universal proof.[1]

Four terms are easy: add $1/n$ to itself four times. Requiring just three turns that trivial construction into an open question about whole-number denominators. A proof would show that the shorter representation always exists; a counterexample would reveal where this simple compression fails.[1]

The problem isolates how an integer constraint can obstruct shortening a rational decomposition. Its target is uniform in $n$, even though the denominators are allowed to change with each input. A solution would explain whether special-case constructions can always be extended to a three-term representation, or whether some denominator has a genuine obstruction.[1]