Erdős Problem #303

FORMAL RESTATEMENT

Erdős #303; a self-contained finite Ramsey reconstruction, 2026-09-24. Prove that every finite coloring of the positive integers has three distinct same-colored positive integers $a,b,c$ with \[ \frac1a=\frac1b+\frac1c. \] This also answers the formulation with a coloring of all integers, by restricting it to the positive integers.

QUICK LITERATURE AND CONTEXT CHECK

Brown and Rödl proved the affirmative result as part of a general transfer principle for homogeneous equations and their reciprocal forms. The following direct proof uses the same reciprocal-transfer idea but supplies the needed finite Ramsey argument explicitly. Powers of two ensure distinctness, an issue not settled by the usual Schur equation alone.

OBJECTIVE AND ATTACK PLAN

Color the edges between powers of two by the color of a reciprocal denominator associated with their difference. A monochromatic triangle gives three differences satisfying $x+y=z$. Clearing their reciprocals by a common multiple turns this into the desired equation.

WORK

A finite monochromatic-triangle lemma. If the edges of a complete graph are colored with $r\geq1$ colors, then a graph on \[ M=1+r+r^2+\cdots+r^{r+1} \tag{1} \] vertices contains a monochromatic triangle. Here is a direct proof, including the finite bound. Define $T_1=1$ and $T_{j+1}=1+rT_j$, so $T_{r+2}=M$. From a set of at least $T_j$ vertices, choose one vertex. Its edges to the others have some color occurring at least $T_{j-1}$ times, because there are at least $rT_{j-1}$ remaining vertices. Retain that color class of neighbors.

Iterating produces distinct vertices $v_1,\ldots,v_{r+2}$ such that, for each $i\leq r+1$, all edges from $v_i$ to the later chosen vertices have one color $c_i$. Two of $c_1,\ldots,c_{r+1}$ coincide, say $c_i=c_j$ with $i<j$. Choose any $k>j$. Then the three edges $v_iv_j$, $v_iv_k$, and $v_jv_k$ all have that color, proving the lemma.

Transfer to unit fractions. Given an $r$-coloring $\chi$ of the positive integers, use the $M$ integer labels $0,1,\ldots,M-1$ as vertices. Let \[ L=\operatorname{lcm}(1,2,\ldots,2^M). \] For $i<j$, the positive difference $2^j-2^i$ is below $2^M$ and hence divides $L$. Color the edge $\{i,j\}$ by \[ \chi\!\left(\frac L{2^j-2^i}\right). \tag{2} \] By the lemma there are indices $i<j<k$ whose three edges have the same color. Set \[ x=2^j-2^i,\qquad y=2^k-2^j,\qquad z=2^k-2^i=x+y. \] All three numbers divide $L$. Thus \[ a=L/z,\qquad b=L/x,\qquad c=L/y \tag{3} \] are positive integers and, by $2$, have the same color. Furthermore \[ \frac1a=\frac zL=\frac{x+y}L=\frac1b+\frac1c. \]

Finally, $z>x,y$ implies $a<b,c$. Also \[ v_2(x)=i,\qquad v_2(y)=j, \] because the factors $2^{j-i}-1$ and $2^{k-j}-1$ are odd. Since $i<j$, we have $x\neq y$, and therefore $b\neq c$. This verifies pairwise distinctness in full.

VERIFICATION AND SCOPE

All arguments are finite and supply an explicit, though extremely large, interval bound: every denominator in $3$ is at most $L$. The proof imports no Ramsey theorem without proof. The powers of two eliminate the degenerate possibility $b=c$, and the common multiple ensures that reciprocal transfer yields integers. The result and transfer method are attributed to the established literature.

SOURCES

• Current problem: https://www.erdosproblems.com/303.
• T. C. Brown and V. Rödl, Monochromatic solutions to equations with unit fractions, Bulletin of the Australian Mathematical Society 43 $1991$, 387--392: https://doi.org/10.1017/S0004972700029221.

CONCLUSION

Attempt status: solved. The finite Ramsey construction gives three distinct same-colored positive integers satisfying the required reciprocal equation. Completion rate estimate: 100%; the proof is self-contained.