The Erdős-Straus Ladder: Where the Easy Cases End

The Erdős–Straus conjecture asks for three unit fractions whose sum is $4/n$, for every $n\\ge2$. Nearly every congruence class yields to a familiar formula. The difficulty is concentrated in the class $n\\equiv1\\pmod4$ with $3 n$, but even there the numbers do not behave alike: many give up a decomposition immediately, while a smaller set seems to demand a real search. This paper explains that difference. For $n\\equiv1\\pmod4$ and $r\\equiv3\\pmod4$, set $a=(n+r)/4$ and $M=na$. Choosing $1/a$ leaves $r/M$. That remainder splits into two unit fractions precisely when a divisor $u$ of $M^2$ satisfies $u-M r$. The problem becomes tangible: at each rung $r$, look for a divisor in one prescribed residue class. Sometimes that divisor is already visible in $n$, $n+1$, or $n+4$; sometimes the factorisation of $M$ must do more work. We show that the monomial divisors $n^ia^j$ yield exactly the three direct families arising from $r n$, $r n+1$, and $r n+4$. We then follow the survivors. Between $5$ and $200000$, these families and the case $n\\equiv5\\pmod8$ cover $32062$ of the $33333$ residual values. The remaining values lie in $n\\equiv124$. Divisor shifts and multiplicative descent reduce the list to $624$ cases, and a rung-by-rung divisor test gives an explicit decomposition for each. The unresolved question is now a clean one: can the first successful rung be bounded as a function of $n$?