gpt pro 5.4 attempt on Math-5526

Erdős Problem #242 (Erdős--Straus)

1) ROUND-2 OBJECTIVE

I am pursuing (C) obstruction/correction. Round-1 reduced the problem only to the classes $n\equiv 1,5 \ (\mathrm{mod}\ 6)$. The most promising Round-2 direction is to squeeze the hard set much further by proving explicit new decomposition formulas and then converting them into strong necessary conditions for any counterexample.

2) Round-1 FOUNDATION USED

I use the following Round-1 facts.

1. Scaling lemma. If \[ \frac{4}{n}=\frac1x+\frac1y+\frac1z, \] then for every positive integer $m$, \[ \frac{4}{mn}=\frac1{mx}+\frac1{my}+\frac1{mz}. \]
2. Prime reduction. If a counterexample exists, then some prime counterexample exists.
3. Round-1 explicit decompositions. All $n$ divisible by $2$ are solvable; all $n$ divisible by $3$ are solvable.

3) NEW INSIGHT / TOOL (ROUND-2)

There are three genuinely new formulas in this round.

1. An Obláth-type divisor criterion: if $n+1$ has a divisor congruent to $3 \ (\mathrm{mod}\ 4)$, then $4/n$ has a three-term unit-fraction decomposition.
2. A direct formula for every $n\equiv 3 \ (\mathrm{mod}\ 4)$.
3. A direct formula for every $n\equiv 5 \ (\mathrm{mod}\ 8)$.

Together these shrink the only possible counterexamples all the way down to \[ n\equiv 1 \ (\mathrm{mod}\ 24), \] and in fact force every prime divisor of a counterexample to lie in the same residue class.

4) ATTACK PLAN (ROUND-2)

The exact Round-1 gap was the large remaining set of odd integers coprime to $6$. I will now prove three new infinite families of explicit solutions and then intersect their complements. The target claims are:

1. if $n+1$ has a divisor $\equiv 3 \ (\mathrm{mod}\ 4)$, then $n$ is not a counterexample;
2. if $n\equiv 3 \ (\mathrm{mod}\ 4)$, then $n$ is not a counterexample;
3. if $n\equiv 5 \ (\mathrm{mod}\ 8)$, then $n$ is not a counterexample;
4. therefore any counterexample must satisfy $n\equiv 1 \ (\mathrm{mod}\ 24)$;
5. by prime reduction, every prime divisor of any counterexample must also be congruent to $1 \ (\mathrm{mod}\ 24)$.

This is a genuine strengthening of Round-1 because it cuts the hard set from two residue classes modulo $6$ down to one residue class modulo $24$, plus extra multiplicative restrictions.

5) WORK (ROUND-2)

Proposition 5.1 (Obláth-type divisor criterion).

Assume $n>2$ is odd and \[ n+1=m(4k-1) \] with integers $k\ge 1$ and $m\ge 2$. Then \[ \frac{4}{n}=\frac{1}{km}+\frac{1}{kn}+\frac{1}{kmn}, \] with three distinct positive denominators.

Proof. Compute: \[ \frac{1}{km}+\frac{1}{kn}+\frac{1}{kmn} =\frac{n+m+1}{kmn}. \] Since $n=(4k-1)m-1$, we have \[ n+m+1=((4k-1)m-1)+m+1=4km, \] so the right-hand side equals $4/n$.

It remains to check distinctness. Because $n>2$ is odd and $n+1=m(4k-1)$ with $4k-1$ odd, the factor $m$ is even, hence $m\ge 2$. Also \[ n=(4k-1)m-1>m, \] so $km<kn$, and since $m\ge 2$ we also have $kn<kmn$. Thus the three denominators are distinct.

Corollary 5.2.

If $n>2$ and $n+1$ has a divisor congruent to $3 \ (\mathrm{mod}\ 4)$, then $n$ satisfies the Erdős--Straus conclusion.

Proof. If $n$ is even, Round-1 already settles it. If $n$ is odd, write the divisor as $4k-1$ and put $m=(n+1)/(4k-1)$. Then $m\ge 2$ as above, and Proposition 5.1 applies.

Proposition 5.3 (all $n\equiv 3 \ (\mathrm{mod}\ 4)$).

If $n=4k+3$ with $k\ge 0$, then \[ \frac{4}{n}=\frac{1}{k+1}+\frac{1}{M+1}+\frac{1}{M(M+1)}, \qquad M:=(k+1)n. \] Hence every $n\equiv 3 \ (\mathrm{mod}\ 4)$ is solvable.

Proof. First, \[ \frac{1}{k+1}+\frac{1}{M} =\frac{n+1}{(k+1)n} =\frac{4(k+1)}{(k+1)n} =\frac{4}{n}. \] Now split $1/M$ using the identity \[ \frac{1}{M}=\frac{1}{M+1}+\frac{1}{M(M+1)}. \] This gives the claimed decomposition.

Distinctness is immediate: $k+1<M+1<M(M+1)$ because $M=(k+1)n\ge 3(k+1)$.

Proposition 5.4 (all $n\equiv 5 \ (\mathrm{mod}\ 8)$).

If $n=8k+5$ with $k\ge 0$, then \[ \frac{4}{n} =\frac{1}{2(k+1)}+\frac{1}{(k+1)n}+\frac{1}{2(k+1)n}. \] Hence every $n\equiv 5 \ (\mathrm{mod}\ 8)$ is solvable.

Proof. Let $A=(k+1)n$. Then \[ \frac{1}{A}+\frac{1}{2A}=\frac{3}{2A}, \] so \[ \frac{1}{2(k+1)}+\frac{1}{A}+\frac{1}{2A} =\frac{n+3}{2(k+1)n}. \] Because $n=8k+5$, we have $n+3=8(k+1)$, hence the last expression is $4/n$. Distinctness is clear because \[ 2(k+1)<(k+1)n<2(k+1)n \] for every $n>2$.

Theorem 5.5 (sharp residue-class reduction obtained in Round-2).

If $n>2$ is a counterexample to the Erdős--Straus conjecture, then \[ n\equiv 1 \ (\mathrm{mod}\ 24). \] Moreover:

1. every prime divisor of $n$ is congruent to $1 \ (\mathrm{mod}\ 24)$;
2. every odd prime divisor of $n+1$ is congruent to $1 \ (\mathrm{mod}\ 4)$.

Proof. By Round-1, a counterexample cannot be even and cannot be divisible by $3$. So $n$ is odd.

By Proposition 5.3, a counterexample cannot satisfy $n\equiv 3 \ (\mathrm{mod}\ 4)$. Hence \[ n\equiv 1 \ (\mathrm{mod}\ 4). \] By Proposition 5.4, a counterexample cannot satisfy $n\equiv 5 \ (\mathrm{mod}\ 8)$. Since an odd number congruent to $1 \ (\mathrm{mod}\ 4)$ is either $1$ or $5 \ (\mathrm{mod}\ 8)$, we deduce \[ n\equiv 1 \ (\mathrm{mod}\ 8). \] By Corollary 5.2, a counterexample cannot satisfy $n\equiv 2 \ (\mathrm{mod}\ 3)$, because then $3\mid (n+1)$ and $3\equiv 3 \ (\mathrm{mod}\ 4)$. Since $3$ does not divide $n$ already, this forces \[ n\equiv 1 \ (\mathrm{mod}\ 3). \] Combining $n\equiv 1 \ (\mathrm{mod}\ 8)$ and $n\equiv 1 \ (\mathrm{mod}\ 3)$ yields \[ n\equiv 1 \ (\mathrm{mod}\ 24), \] as claimed.

For the prime-factor statement, let $p$ be any prime divisor of a counterexample $n$. By the Round-1 scaling lemma, if $p$ were solvable then $n$ would also be solvable. So $p$ is itself a counterexample. Applying the residue-class conclusion to $p$ gives \[ p\equiv 1 \ (\mathrm{mod}\ 24). \] Finally, if an odd prime divisor $r$ of $n+1$ satisfied $r\equiv 3 \ (\mathrm{mod}\ 4)$, then Corollary 5.2 would solve $n$, contradiction. Therefore every odd prime divisor of $n+1$ is $1 \ (\mathrm{mod}\ 4)$.

6) ADVERSARIAL VERIFICATION

1. Distinctness in Proposition 5.1. The only delicate case is $m=1$, which would give repeated denominators. But when $n$ is odd this cannot happen: $n+1$ is even while $4k-1$ is odd, so $m$ must be even.
2. Does Proposition 5.4 really add something new? Yes. The residue class $13 \ (\mathrm{mod}\ 24)$ is not covered by the Round-1 even/$3\mid n$ cases and is not covered by Proposition 5.3, but it is covered by Proposition 5.4.
3. Did I use prime reduction correctly? Yes. If a prime divisor $p$ of $n$ were solvable, scaling by $n/p$ would solve $n$.
4. Any missed residue class? The surviving classes modulo $24$ after Round-1 plus Propositions 5.3 and 5.4 are $1$ and $17$. Corollary 5.2 eliminates $17$ because then $3\mid (n+1)$. So only $1 \ (\mathrm{mod}\ 24)$ remains.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

Round-2 strengthens the necessary conditions on any counterexample from \[ n\equiv 1,5 \ (\mathrm{mod}\ 6) \] all the way down to \[ n\equiv 1 \ (\mathrm{mod}\ 24), \] with two further multiplicative obstructions:

1. every prime factor of $n$ must itself be $1 \ (\mathrm{mod}\ 24)$;
2. every odd prime factor of $n+1$ must be $1 \ (\mathrm{mod}\ 4)$.

This is a genuine gap-closure step, but it does not yet resolve the conjecture.

8) COMPLETION ESTIMATE

COMPLETION: 58%

9) REFERENCES

1. T. F. Bloom, Erdős Problem #242, current version page on erdosproblems.com, accessed 2026-03-07.
2. S. Mihnea and D. C. Bogdan, Further verification and empirical evidence for the Erdős--Straus conjecture, arXiv:2509.00128 (checked for current computational status, though not used in the proofs above).