The Erdős–Straus Conjecture: Egyptian Fractions Nobody Can Finish

Introduction

The scribes of ancient Egypt did not write $\frac{3}{4}$ . They wrote $\frac12 + \frac14$ . Apart from a few special symbols, every fraction was expressed as a sum of unit fractions — fractions with numerator 1. The Rhind Mathematical Papyrus (copied by the scribe Ahmes around 1550 BC) opens with a table that rewrites $\frac{2}{n}$ this way for every odd $n$ from 5 to 101, for example $\frac{2}{5} = \frac13 + \frac{1}{15}$ .

Such sums are still called Egyptian fractions, and every positive fraction has one. The simplest recipe is the greedy algorithm, described by Fibonacci in Liber Abaci (1202) and analysed by J. J. Sylvester in 1880: take the largest unit fraction that fits, subtract it, and repeat.

greedy(p/q):              # 0 < p/q < 1
  terms = []
  while p > 0:
    d = ceil(q / p)       # largest 1/d with 1/d ≤ p/q
    terms.append(d)
    p, q = p*d - q, q*d   # p/q − 1/d, then reduce with gcd
  return terms

Each step strictly lowers the numerator, so it always stops — but it can be wasteful. Greedy turns $\frac{4}{17}$ into four terms with a seven-digit denominator, $\frac{4}{17} = \frac15 + \frac{1}{29} + \frac{1}{1233} + \frac{1}{3039345}$ , although three are enough: $\frac{4}{17} = \frac15 + \frac{1}{30} + \frac{1}{510}$ .

That raises a natural question. How many unit fractions does $\frac{4}{n}$ really need? Two are not always enough, but in 1948 Paul Erdős and Ernst G. Straus conjectured that three always are:

\frac{4}{n} = \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \quad \text{has a solution in positive integers for every } n \ge 2.

Examples are easy to find: $\frac45 = \frac12 + \frac14 + \frac{1}{20}$ and $\frac47 = \frac12 + \frac{1}{15} + \frac{1}{210}$ . A general proof has never been found.

Split 4/n Yourself

Type a number $n$ and press Find solutions. The demo first checks whether $n$ is in a residue class where a ready-made identity gives the answer straight away. Then it runs a real search: it tries each candidate $x$ in the window $\frac{n}{4} < x \le \frac{3n}{4}$ and, for each one, finds every way to split what is left into $\frac1y + \frac1z$ . Every triple is checked in exact integer arithmetic.

<!-- Erdős–Straus explorer: solve 4/n = 1/x + 1/y + 1/z, chart how many solutions each n has, and compare with the greedy algorithm. -->
<p class="hint">The conjecture says that for every n ≥ 2 the fraction 4/n is a sum of three unit fractions 1/x + 1/y + 1/z. Type any n: the page tells you whether a <b>modular identity</b> solves it instantly or whether it lands in one of the six <b>hard classes mod 840</b> where only search works, then finds and checks the triples.</p>
<div class="row">
  <label for="n">n =</label>
  <input id="n" type="number" min="2" max="10000000" value="1009" aria-label="Whole number n between 2 and 10 million">
  <button id="find" type="button">Find solutions</button>
</div>
<div class="btns">
  <button id="rand" type="button" class="ghost">Random prime</button>
  <button id="hard" type="button" class="ghost">Hard class: 1009</button>
  <button id="easy" type="button" class="ghost">Easy: 1003</button>
</div>
<div class="out" id="out" aria-live="polite"></div>
<h4>How many solutions does each n have?</h4>
<button id="chartbtn" type="button">Count solutions for n ≤ 300</button>
<canvas id="cv" width="600" height="240" aria-label="Bar chart of the number of solutions for each n from 2 to 300"></canvas>
<p class="legend">Dark bars: primes, the only cases that really matter. Light bars: composites, which inherit extra solutions from their factors. Vertical axis is logarithmic. Tap a bar to load that n.</p>
<div class="out" id="chartout" aria-live="polite"></div>
<h4>For comparison, the greedy Egyptian-fraction algorithm</h4>
<div class="row">
  <input id="frac" type="text" value="4/17" spellcheck="false" autocomplete="off" aria-label="A fraction p/q such as 4/17">
  <button id="greedy" type="button">Expand greedily</button>
</div>
<div class="out" id="gout" aria-live="polite"></div>

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const i18nFmt = (s, v) => s.replace(/%(\w+)%/g, (_, k) => String(v[k] ?? ''));
const gcd = (a, b) => { while (b) { const t = a % b; a = b; b = t; } return a; };
const bgcd = (a, b) => { while (b) { const t = a % b; a = b; b = t; } return a; };
// Trial-division factorisation; fine for numbers up to ten million.
function factor(m, f) {
  for (let p = 2; p * p <= m; p += (p === 2 ? 1 : 2)) {
    while (m % p === 0) { f.set(p, (f.get(p) || 0) + 1); m /= p; }
  }
  if (m > 1) f.set(m, (f.get(m) || 0) + 1);
  return f;
}
const isPrime = (m) => { if (m < 2) return false; const f = factor(m, new Map()); return f.size === 1 && f.get(m) === 1; };
// For a fixed x, the rest a/b = 4/n − 1/x splits as 1/y + 1/z exactly when (ay − b)(az − b) = b², so we walk the divisors d ≤ b of b².
function pairsFor(n, x, limit, out) {
  let a = 4 * x - n, b = n * x;
  const g = gcd(a, b); a /= g; b /= g;
  const f = new Map(); factor(n, f); factor(x, f);
  const pe = [];
  for (const p of f.keys()) { let e = 0, t = b; while (t % p === 0) { t /= p; e++; } if (e) pe.push([p, 2 * e]); }
  let divs = [1];
  for (const [p, e] of pe) {
    const nd = [];
    for (const d of divs) { let v = d; for (let i = 0; i <= e; i++) { if (v > b) break; nd.push(v); v *= p; } }
    divs = nd;
  }
  divs.sort((u, v) => u - v);
  const B = BigInt(b), A = BigInt(a);
  for (const d of divs) {
    if (d > b) break;
    if ((d + b) % a !== 0) continue;
    const y = (d + b) / a;
    if (y < x) continue;
    out.push([BigInt(x), BigInt(y), (B * B / BigInt(d) + B) / A]);
    if (out.length >= limit) return;
  }
}
// The smallest denominator x satisfies n/4 < x ≤ 3n/4; try each x in that window.
function solutions(n, limit) {
  const out = [];
  const lo = Math.floor(n / 4) + 1, hi = Math.floor(3 * n / 4);
  let tried = 0;
  for (let x = lo; x <= hi && out.length < limit; x++) { tried++; pairsFor(n, x, limit, out); }
  return { out, tried, lo: lo - 1, hi };
}
const checks = (n, s) => { const N = BigInt(n), [x, y, z] = s; return 4n * x * y * z === N * (y * z + x * z + x * y); };
const OPEN = [1, 121, 169, 289, 361, 529];
// Modular identities: when n falls in one of these residue classes, a formula hands us a solution with no search at all.
function identity(n) {
  const B = BigInt(n);
  if (n % 2 === 0) return { key: 'mod2', sol: [B / 2n, B, B] };
  if (n % 4 === 3) { const a = (B + 1n) / 4n; return { key: 'mod4', sol: [a, 2n * a * B, 2n * a * B] }; }
  if (n % 3 === 2) { const b = (B + 1n) / 3n; return { key: 'mod3', sol: [b, B, b * B] }; }
  if (n % 8 === 5) { const c = (B + 3n) / 8n; return { key: 'mod8', sol: [2n * c, c * B, 2n * c * B] }; }
  if (n % 5 === 2 || n % 5 === 3) return { key: 'mod5', r: n % 5 };
  if (n % 7 === 3 || n % 7 === 5 || n % 7 === 6) return { key: 'mod7', r: n % 7 };
  if (n % 3 === 0) { const m = B / 3n; return { key: 'mod3z', sol: [m, 4n * m, 12n * m] }; }
  if (n % 5 === 0) { const m = B / 5n; return { key: 'mod5z', sol: [2n * m, 4n * m, 20n * m] }; }
  if (n % 7 === 0) { const m = B / 7n; return { key: 'mod7z', sol: [2n * m, 15n * m, 210n * m] }; }
  for (const p of factor(n, new Map()).keys()) {
    if (p === n) continue;
    const s = identity(p);
    if (s.key !== 'open') { const K = BigInt(n / p); return { key: 'scale', p: p, k: n / p, sol: s.sol ? s.sol.map((v) => v * K) : null }; }
  }
  return { key: 'open', r: n % 840 };
}
const CLS = {
  mod2: "n is even, so 4/n = 1/(n/2) + 1/n + 1/n.", mod4: "n ≡ 3 (mod 4): with a = (n + 1)/4, 4/n = 1/a + 1/(2an) + 1/(2an).", mod3: "n ≡ 2 (mod 3): with b = (n + 1)/3, 4/n = 1/b + 1/n + 1/(bn).", mod8: "n ≡ 5 (mod 8): with c = (n + 3)/8, 4/n = 1/(2c) + 1/(cn) + 1/(2cn).",
  mod5: "n ≡ %r% (mod 5): one of Mordell’s polynomial identities covers this class; the search below produces the triple.", mod7: "n ≡ %r% (mod 7): one of Mordell’s polynomial identities covers this class; the search below produces the triple.", mod3z: "n is a multiple of 3: scale 4/3 = 1/1 + 1/4 + 1/12 by n/3.", mod5z: "n is a multiple of 5: scale 4/5 = 1/2 + 1/4 + 1/20 by n/5.",
  mod7z: "n is a multiple of 7: scale 4/7 = 1/2 + 1/15 + 1/210 by n/7.", scale: "n = %p% × %k%: solve 4/%p% with an identity and multiply every denominator by %k%.", open: "n ≡ %r% (mod 840): one of the six hard classes (1, 121, 169, 289, 361, 529). No known identity covers it — only a search finds the triple."
};
const tri = (s) => '1/' + s[0] + ' + 1/' + s[1] + ' + 1/' + s[2];

// Interface: classify n, show the identity (if any), then search and list the triples.
const $ = (id) => document.getElementById(id);
function addLine(box, cls, text) { const d = document.createElement('div'); d.className = 'line ' + cls; d.textContent = text; box.appendChild(d); return d; }
function run(n) {
  const out = $('out'); out.innerHTML = '';
  if (!Number.isInteger(n) || n < 2 || n > 10000000) { addLine(out, 'bad', "Enter a whole number from 2 to 10,000,000."); return; }
  $('n').value = n;
  const id = identity(n);
  addLine(out, 'lbl', i18nFmt("n = %n% (%p%), n mod 840 = %r%", { n: n, r: n % 840, p: isPrime(n) ? "prime" : "composite" }));
  addLine(out, id.key === 'open' ? 'warn' : 'ok', i18nFmt(CLS[id.key], { r: id.r, p: id.p, k: id.k }));
  if (id.sol) {
    addLine(out, 'mono', '4/' + n + ' = ' + tri(id.sol) + (checks(n, id.sol) ? '  ✓' : '  ✗'));
  }
  const t0 = performance.now();
  const big = n > 5000;
  const res = solutions(n, big ? 8 : 100000);
  const ms = Math.max(0, performance.now() - t0).toFixed(1);
  if (!res.out.length) { addLine(out, 'bad', "No solution found. That would be a counterexample to the conjecture — it has never happened."); return; }
  const first = solutions(n, 1);
  addLine(out, 'lbl', i18nFmt("Direct search: the first triple appeared after trying %t% value(s) of x in the window (%lo%, %hi%].", { t: first.tried, lo: first.lo, hi: first.hi }));
  addLine(out, 'mono', '4/' + n + ' = ' + tri(first.out[0]) + (checks(n, first.out[0]) ? '  ✓' : '  ✗'));
  addLine(out, 'lbl', i18nFmt(big ? "n is large, so the list stops after the first %c% solutions (%ms% ms):" : "All %c% solutions with x ≤ y ≤ z (%ms% ms):", { c: res.out.length, ms: ms }));
  const box = document.createElement('div'); box.className = 'sols'; out.appendChild(box);
  for (const s of res.out.slice(0, 200)) addLine(box, 'mono', '(' + s.join(', ') + ')');
}
$('find').addEventListener('click', () => run(Number($('n').value)));
$('n').addEventListener('keydown', (e) => { if (e.key === 'Enter') run(Number($('n').value)); });
$('hard').addEventListener('click', () => run(1009));
$('easy').addEventListener('click', () => run(1003));
$('rand').addEventListener('click', () => {
  // Half the time pick a random prime from one of the six hard classes mod 840, otherwise any random prime.
  const want = Math.random() < 0.5;
  for (let k = 0; k < 5000; k++) {
    const p = 2 + Math.floor(Math.random() * 9999990);
    if (!isPrime(p)) continue;
    if (want && !OPEN.includes(p % 840)) continue;
    run(p); return;
  }
  run(1009);
});

// Solution counts f(n) for n ≤ 300, computed in small chunks so the page stays responsive; log-scale bars.
const counts = [];
function dark() { return window.matchMedia && window.matchMedia('(prefers-color-scheme: dark)').matches; }
function drawChart(sel) {
  const cv = $('cv'), ctx = cv.getContext('2d'), W = cv.width, H = cv.height, D = dark();
  ctx.fillStyle = D ? '#16202b' : '#f7f9fb'; ctx.fillRect(0, 0, W, H);
  const L = 34, Bm = 22, T = 10, R = 6;
  const maxC = Math.max(2, ...counts.filter(Boolean));
  const ly = (c) => H - Bm - (Math.log10(c) / Math.log10(maxC)) * (H - Bm - T);
  ctx.strokeStyle = D ? '#2c3b4b' : '#dde3ea'; ctx.fillStyle = D ? '#b4c0cc' : '#556'; ctx.font = '11px system-ui, sans-serif'; ctx.lineWidth = 1;
  for (let c = 1; c <= maxC; c *= 10) { const y = ly(c); ctx.beginPath(); ctx.moveTo(L, y); ctx.lineTo(W - R, y); ctx.stroke(); ctx.fillText(String(c), 4, y + 4); }
  for (let n = 50; n <= 300; n += 50) ctx.fillText(String(n), L + (n - 2) / 299 * (W - L - R) - 8, H - 6);
  const bw = (W - L - R) / 299;
  for (let n = 2; n <= 300; n++) {
    const c = counts[n]; if (!c) continue;
    const pr = isPrime(n);
    ctx.fillStyle = n === sel ? (D ? '#f0c060' : '#d08700') : pr ? (D ? '#7fb2ff' : '#1d3557') : (D ? '#3a5068' : '#b9c7d8');
    const x = L + (n - 2) * bw, y = Math.min(ly(Math.max(c, 1)), H - Bm - 2);
    ctx.fillRect(x, y, Math.max(1, bw - 0.4), H - Bm - y);
  }
}
$('chartbtn').addEventListener('click', () => {
  const btn = $('chartbtn'), co = $('chartout'); btn.disabled = true;
  let n = 2;
  function step() {
    const until = Math.min(300, n + 12);
    for (; n <= until; n++) counts[n] = solutions(n, 100000).out.length;
    co.textContent = i18nFmt("Counting… n = %n% of 300", { n: n - 1 });
    drawChart(0);
    if (n <= 300) { setTimeout(step, 0); return; }
    let bp = 0, bm = 2;
    for (let k = 2; k <= 300; k++) { if (isPrime(k) && (!bp || counts[k] < counts[bp])) bp = k; if (counts[k] > counts[bm]) bm = k; }
    co.textContent = i18nFmt("Done. Fewest solutions among primes: f(%p%) = %c%. Most solutions overall: f(%m%) = %mc%.", { p: bp, c: counts[bp], m: bm, mc: counts[bm] });
    btn.disabled = false;
  }
  step();
});
$('cv').addEventListener('click', (e) => {
  if (counts.length < 300) return;
  const r = $('cv').getBoundingClientRect(), W = $('cv').width;
  const px = (e.clientX - r.left) * W / r.width;
  const n = Math.floor(2 + (px - 34) / ((W - 40) / 299));
  if (n >= 2 && n <= 300) { drawChart(n); run(n); }
});

// Greedy (Fibonacci–Sylvester): always subtract the largest unit fraction that fits.
function greedy(p, q) {
  let P = BigInt(p), Q = BigInt(q); const terms = [];
  while (P > 0n && terms.length < 10) {
    const d = (Q + P - 1n) / P; terms.push(d);
    P = P * d - Q; Q = Q * d;
    if (P) { const g = bgcd(P, Q); P /= g; Q /= g; }
  }
  return { terms, done: P === 0n };
}
const short = (b) => { const s = b.toString(); return s.length <= 24 ? s : s.slice(0, 10) + '…' + i18nFmt("(%d% digits)", { d: s.length }); };
$('greedy').addEventListener('click', () => {
  const g = $('gout'); g.innerHTML = '';
  const m = /^\s*(\d+)\s*\/\s*(\d+)\s*$/.exec($('frac').value);
  const p = m ? Number(m[1]) : 0, q = m ? Number(m[2]) : 0;
  if (!m || p < 1 || p >= q || q > 1000000000) { addLine(g, 'bad', "Type a fraction like 4/17 with 0 < p < q ≤ 1,000,000,000."); return; }
  const r = greedy(p, q);
  addLine(g, 'mono', p + '/' + q + ' = ' + r.terms.map((d) => '1/' + short(d)).join(' + ') + (r.done ? '' : ' + …'));
  addLine(g, r.done ? 'ok' : 'warn', i18nFmt(r.done ? "The greedy algorithm needed %k% term(s)." : "The greedy algorithm is still going after %k% terms and the denominators are exploding.", { k: r.terms.length }));
  const k = gcd(p, q);
  if (p / k === 4 && q / k <= 10000000) {
    const s = solutions(q / k, 1).out[0];
    if (s) addLine(g, 'lbl', i18nFmt("Yet a three-term expansion exists: 4/%q% = %t%.", { q: q / k, t: tri(s) }));
  }
});
run(1009);

<!-- Erdős–Straus explorer: solve 4/n = 1/x + 1/y + 1/z, chart how many solutions each n has, and compare with the greedy algorithm. -->
<p class="hint">The conjecture says that for every n ≥ 2 the fraction 4/n is a sum of three unit fractions 1/x + 1/y + 1/z. Type any n: the page tells you whether a <b>modular identity</b> solves it instantly or whether it lands in one of the six <b>hard classes mod 840</b> where only search works, then finds and checks the triples.</p>
<div class="row">
  <label for="n">n =</label>
  <input id="n" type="number" min="2" max="10000000" value="1009" aria-label="Whole number n between 2 and 10 million">
  <button id="find" type="button">Find solutions</button>
</div>
<div class="btns">
  <button id="rand" type="button" class="ghost">Random prime</button>
  <button id="hard" type="button" class="ghost">Hard class: 1009</button>
  <button id="easy" type="button" class="ghost">Easy: 1003</button>
</div>
<div class="out" id="out" aria-live="polite"></div>
<h4>How many solutions does each n have?</h4>
<button id="chartbtn" type="button">Count solutions for n ≤ 300</button>
<canvas id="cv" width="600" height="240" aria-label="Bar chart of the number of solutions for each n from 2 to 300"></canvas>
<p class="legend">Dark bars: primes, the only cases that really matter. Light bars: composites, which inherit extra solutions from their factors. Vertical axis is logarithmic. Tap a bar to load that n.</p>
<div class="out" id="chartout" aria-live="polite"></div>
<h4>For comparison, the greedy Egyptian-fraction algorithm</h4>
<div class="row">
  <input id="frac" type="text" value="4/17" spellcheck="false" autocomplete="off" aria-label="A fraction p/q such as 4/17">
  <button id="greedy" type="button">Expand greedily</button>
</div>
<div class="out" id="gout" aria-live="polite"></div>

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.line.bad { color: #c92f3c; font-weight: 600; }
.line.warn { color: #8a5a00; font-weight: 600; }
.sols { max-height: 180px; overflow: auto; }
@media (prefers-color-scheme: dark) {
  body { color: #e3e8ee; background: #0f1720; }
  .hint, .legend { color: #b4c0cc; }
  h4 { color: #9fc4ff; }
  input { background: #16202b; color: #e3e8ee; border-color: #3a4a5c; }
  canvas { border-color: #2c3b4b; }
  button { background: #7fb2ff; border-color: #7fb2ff; color: #0f1720; }
  button.ghost { background: transparent; color: #7fb2ff; }
  .line.lbl { color: #9fc4ff; }
  .line.mono { background: #1b2733; }
  .line.ok { color: #5fd08a; }
  .line.bad { color: #ff7b86; }
  .line.warn { color: #f0c060; }
}

const i18nFmt = (s, v) => s.replace(/%(\w+)%/g, (_, k) => String(v[k] ?? ''));
const gcd = (a, b) => { while (b) { const t = a % b; a = b; b = t; } return a; };
const bgcd = (a, b) => { while (b) { const t = a % b; a = b; b = t; } return a; };
// Trial-division factorisation; fine for numbers up to ten million.
function factor(m, f) {
  for (let p = 2; p * p <= m; p += (p === 2 ? 1 : 2)) {
    while (m % p === 0) { f.set(p, (f.get(p) || 0) + 1); m /= p; }
  }
  if (m > 1) f.set(m, (f.get(m) || 0) + 1);
  return f;
}
const isPrime = (m) => { if (m < 2) return false; const f = factor(m, new Map()); return f.size === 1 && f.get(m) === 1; };
// For a fixed x, the rest a/b = 4/n − 1/x splits as 1/y + 1/z exactly when (ay − b)(az − b) = b², so we walk the divisors d ≤ b of b².
function pairsFor(n, x, limit, out) {
  let a = 4 * x - n, b = n * x;
  const g = gcd(a, b); a /= g; b /= g;
  const f = new Map(); factor(n, f); factor(x, f);
  const pe = [];
  for (const p of f.keys()) { let e = 0, t = b; while (t % p === 0) { t /= p; e++; } if (e) pe.push([p, 2 * e]); }
  let divs = [1];
  for (const [p, e] of pe) {
    const nd = [];
    for (const d of divs) { let v = d; for (let i = 0; i <= e; i++) { if (v > b) break; nd.push(v); v *= p; } }
    divs = nd;
  }
  divs.sort((u, v) => u - v);
  const B = BigInt(b), A = BigInt(a);
  for (const d of divs) {
    if (d > b) break;
    if ((d + b) % a !== 0) continue;
    const y = (d + b) / a;
    if (y < x) continue;
    out.push([BigInt(x), BigInt(y), (B * B / BigInt(d) + B) / A]);
    if (out.length >= limit) return;
  }
}
// The smallest denominator x satisfies n/4 < x ≤ 3n/4; try each x in that window.
function solutions(n, limit) {
  const out = [];
  const lo = Math.floor(n / 4) + 1, hi = Math.floor(3 * n / 4);
  let tried = 0;
  for (let x = lo; x <= hi && out.length < limit; x++) { tried++; pairsFor(n, x, limit, out); }
  return { out, tried, lo: lo - 1, hi };
}
const checks = (n, s) => { const N = BigInt(n), [x, y, z] = s; return 4n * x * y * z === N * (y * z + x * z + x * y); };
const OPEN = [1, 121, 169, 289, 361, 529];
// Modular identities: when n falls in one of these residue classes, a formula hands us a solution with no search at all.
function identity(n) {
  const B = BigInt(n);
  if (n % 2 === 0) return { key: 'mod2', sol: [B / 2n, B, B] };
  if (n % 4 === 3) { const a = (B + 1n) / 4n; return { key: 'mod4', sol: [a, 2n * a * B, 2n * a * B] }; }
  if (n % 3 === 2) { const b = (B + 1n) / 3n; return { key: 'mod3', sol: [b, B, b * B] }; }
  if (n % 8 === 5) { const c = (B + 3n) / 8n; return { key: 'mod8', sol: [2n * c, c * B, 2n * c * B] }; }
  if (n % 5 === 2 || n % 5 === 3) return { key: 'mod5', r: n % 5 };
  if (n % 7 === 3 || n % 7 === 5 || n % 7 === 6) return { key: 'mod7', r: n % 7 };
  if (n % 3 === 0) { const m = B / 3n; return { key: 'mod3z', sol: [m, 4n * m, 12n * m] }; }
  if (n % 5 === 0) { const m = B / 5n; return { key: 'mod5z', sol: [2n * m, 4n * m, 20n * m] }; }
  if (n % 7 === 0) { const m = B / 7n; return { key: 'mod7z', sol: [2n * m, 15n * m, 210n * m] }; }
  for (const p of factor(n, new Map()).keys()) {
    if (p === n) continue;
    const s = identity(p);
    if (s.key !== 'open') { const K = BigInt(n / p); return { key: 'scale', p: p, k: n / p, sol: s.sol ? s.sol.map((v) => v * K) : null }; }
  }
  return { key: 'open', r: n % 840 };
}
const CLS = {
  mod2: "n is even, so 4/n = 1/(n/2) + 1/n + 1/n.", mod4: "n ≡ 3 (mod 4): with a = (n + 1)/4, 4/n = 1/a + 1/(2an) + 1/(2an).", mod3: "n ≡ 2 (mod 3): with b = (n + 1)/3, 4/n = 1/b + 1/n + 1/(bn).", mod8: "n ≡ 5 (mod 8): with c = (n + 3)/8, 4/n = 1/(2c) + 1/(cn) + 1/(2cn).",
  mod5: "n ≡ %r% (mod 5): one of Mordell’s polynomial identities covers this class; the search below produces the triple.", mod7: "n ≡ %r% (mod 7): one of Mordell’s polynomial identities covers this class; the search below produces the triple.", mod3z: "n is a multiple of 3: scale 4/3 = 1/1 + 1/4 + 1/12 by n/3.", mod5z: "n is a multiple of 5: scale 4/5 = 1/2 + 1/4 + 1/20 by n/5.",
  mod7z: "n is a multiple of 7: scale 4/7 = 1/2 + 1/15 + 1/210 by n/7.", scale: "n = %p% × %k%: solve 4/%p% with an identity and multiply every denominator by %k%.", open: "n ≡ %r% (mod 840): one of the six hard classes (1, 121, 169, 289, 361, 529). No known identity covers it — only a search finds the triple."
};
const tri = (s) => '1/' + s[0] + ' + 1/' + s[1] + ' + 1/' + s[2];

// Interface: classify n, show the identity (if any), then search and list the triples.
const $ = (id) => document.getElementById(id);
function addLine(box, cls, text) { const d = document.createElement('div'); d.className = 'line ' + cls; d.textContent = text; box.appendChild(d); return d; }
function run(n) {
  const out = $('out'); out.innerHTML = '';
  if (!Number.isInteger(n) || n < 2 || n > 10000000) { addLine(out, 'bad', "Enter a whole number from 2 to 10,000,000."); return; }
  $('n').value = n;
  const id = identity(n);
  addLine(out, 'lbl', i18nFmt("n = %n% (%p%), n mod 840 = %r%", { n: n, r: n % 840, p: isPrime(n) ? "prime" : "composite" }));
  addLine(out, id.key === 'open' ? 'warn' : 'ok', i18nFmt(CLS[id.key], { r: id.r, p: id.p, k: id.k }));
  if (id.sol) {
    addLine(out, 'mono', '4/' + n + ' = ' + tri(id.sol) + (checks(n, id.sol) ? '  ✓' : '  ✗'));
  }
  const t0 = performance.now();
  const big = n > 5000;
  const res = solutions(n, big ? 8 : 100000);
  const ms = Math.max(0, performance.now() - t0).toFixed(1);
  if (!res.out.length) { addLine(out, 'bad', "No solution found. That would be a counterexample to the conjecture — it has never happened."); return; }
  const first = solutions(n, 1);
  addLine(out, 'lbl', i18nFmt("Direct search: the first triple appeared after trying %t% value(s) of x in the window (%lo%, %hi%].", { t: first.tried, lo: first.lo, hi: first.hi }));
  addLine(out, 'mono', '4/' + n + ' = ' + tri(first.out[0]) + (checks(n, first.out[0]) ? '  ✓' : '  ✗'));
  addLine(out, 'lbl', i18nFmt(big ? "n is large, so the list stops after the first %c% solutions (%ms% ms):" : "All %c% solutions with x ≤ y ≤ z (%ms% ms):", { c: res.out.length, ms: ms }));
  const box = document.createElement('div'); box.className = 'sols'; out.appendChild(box);
  for (const s of res.out.slice(0, 200)) addLine(box, 'mono', '(' + s.join(', ') + ')');
}
$('find').addEventListener('click', () => run(Number($('n').value)));
$('n').addEventListener('keydown', (e) => { if (e.key === 'Enter') run(Number($('n').value)); });
$('hard').addEventListener('click', () => run(1009));
$('easy').addEventListener('click', () => run(1003));
$('rand').addEventListener('click', () => {
  // Half the time pick a random prime from one of the six hard classes mod 840, otherwise any random prime.
  const want = Math.random() < 0.5;
  for (let k = 0; k < 5000; k++) {
    const p = 2 + Math.floor(Math.random() * 9999990);
    if (!isPrime(p)) continue;
    if (want && !OPEN.includes(p % 840)) continue;
    run(p); return;
  }
  run(1009);
});

// Solution counts f(n) for n ≤ 300, computed in small chunks so the page stays responsive; log-scale bars.
const counts = [];
function dark() { return window.matchMedia && window.matchMedia('(prefers-color-scheme: dark)').matches; }
function drawChart(sel) {
  const cv = $('cv'), ctx = cv.getContext('2d'), W = cv.width, H = cv.height, D = dark();
  ctx.fillStyle = D ? '#16202b' : '#f7f9fb'; ctx.fillRect(0, 0, W, H);
  const L = 34, Bm = 22, T = 10, R = 6;
  const maxC = Math.max(2, ...counts.filter(Boolean));
  const ly = (c) => H - Bm - (Math.log10(c) / Math.log10(maxC)) * (H - Bm - T);
  ctx.strokeStyle = D ? '#2c3b4b' : '#dde3ea'; ctx.fillStyle = D ? '#b4c0cc' : '#556'; ctx.font = '11px system-ui, sans-serif'; ctx.lineWidth = 1;
  for (let c = 1; c <= maxC; c *= 10) { const y = ly(c); ctx.beginPath(); ctx.moveTo(L, y); ctx.lineTo(W - R, y); ctx.stroke(); ctx.fillText(String(c), 4, y + 4); }
  for (let n = 50; n <= 300; n += 50) ctx.fillText(String(n), L + (n - 2) / 299 * (W - L - R) - 8, H - 6);
  const bw = (W - L - R) / 299;
  for (let n = 2; n <= 300; n++) {
    const c = counts[n]; if (!c) continue;
    const pr = isPrime(n);
    ctx.fillStyle = n === sel ? (D ? '#f0c060' : '#d08700') : pr ? (D ? '#7fb2ff' : '#1d3557') : (D ? '#3a5068' : '#b9c7d8');
    const x = L + (n - 2) * bw, y = Math.min(ly(Math.max(c, 1)), H - Bm - 2);
    ctx.fillRect(x, y, Math.max(1, bw - 0.4), H - Bm - y);
  }
}
$('chartbtn').addEventListener('click', () => {
  const btn = $('chartbtn'), co = $('chartout'); btn.disabled = true;
  let n = 2;
  function step() {
    const until = Math.min(300, n + 12);
    for (; n <= until; n++) counts[n] = solutions(n, 100000).out.length;
    co.textContent = i18nFmt("Counting… n = %n% of 300", { n: n - 1 });
    drawChart(0);
    if (n <= 300) { setTimeout(step, 0); return; }
    let bp = 0, bm = 2;
    for (let k = 2; k <= 300; k++) { if (isPrime(k) && (!bp || counts[k] < counts[bp])) bp = k; if (counts[k] > counts[bm]) bm = k; }
    co.textContent = i18nFmt("Done. Fewest solutions among primes: f(%p%) = %c%. Most solutions overall: f(%m%) = %mc%.", { p: bp, c: counts[bp], m: bm, mc: counts[bm] });
    btn.disabled = false;
  }
  step();
});
$('cv').addEventListener('click', (e) => {
  if (counts.length < 300) return;
  const r = $('cv').getBoundingClientRect(), W = $('cv').width;
  const px = (e.clientX - r.left) * W / r.width;
  const n = Math.floor(2 + (px - 34) / ((W - 40) / 299));
  if (n >= 2 && n <= 300) { drawChart(n); run(n); }
});

// Greedy (Fibonacci–Sylvester): always subtract the largest unit fraction that fits.
function greedy(p, q) {
  let P = BigInt(p), Q = BigInt(q); const terms = [];
  while (P > 0n && terms.length < 10) {
    const d = (Q + P - 1n) / P; terms.push(d);
    P = P * d - Q; Q = Q * d;
    if (P) { const g = bgcd(P, Q); P /= g; Q /= g; }
  }
  return { terms, done: P === 0n };
}
const short = (b) => { const s = b.toString(); return s.length <= 24 ? s : s.slice(0, 10) + '…' + i18nFmt("(%d% digits)", { d: s.length }); };
$('greedy').addEventListener('click', () => {
  const g = $('gout'); g.innerHTML = '';
  const m = /^\s*(\d+)\s*\/\s*(\d+)\s*$/.exec($('frac').value);
  const p = m ? Number(m[1]) : 0, q = m ? Number(m[2]) : 0;
  if (!m || p < 1 || p >= q || q > 1000000000) { addLine(g, 'bad', "Type a fraction like 4/17 with 0 < p < q ≤ 1,000,000,000."); return; }
  const r = greedy(p, q);
  addLine(g, 'mono', p + '/' + q + ' = ' + r.terms.map((d) => '1/' + short(d)).join(' + ') + (r.done ? '' : ' + …'));
  addLine(g, r.done ? 'ok' : 'warn', i18nFmt(r.done ? "The greedy algorithm needed %k% term(s)." : "The greedy algorithm is still going after %k% terms and the denominators are exploding.", { k: r.terms.length }));
  const k = gcd(p, q);
  if (p / k === 4 && q / k <= 10000000) {
    const s = solutions(q / k, 1).out[0];
    if (s) addLine(g, 'lbl', i18nFmt("Yet a three-term expansion exists: 4/%q% = %t%.", { q: q / k, t: tri(s) }));
  }
});
run(1009);

Result

Console

Try 1009, the smallest prime that none of Mordell's identities covers: the search still finds a triple on its first try. Then press Count solutions for n ≤ 300 and look at the chart. Primes, the only cases that matter, have far fewer solutions than their composite neighbours, yet none of them has zero. Finally, run the greedy algorithm on $\frac{4}{17}$ or $\frac{5}{121}$ and compare its length with the three-term answer.

What Is Actually Known

Only primes matter. If $\frac4p = \frac1x + \frac1y + \frac1z$ , then multiplying every denominator by $k$ gives $\frac{4}{kp} = \frac{1}{kx} + \frac{1}{ky} + \frac{1}{kz}$ . So a solution for a prime $p$ solves every multiple of $p$ . Since even numbers and multiples of 4 are trivial, the conjecture is equivalent to its statement for odd primes.

Identities clear most residue classes. Some classes can be solved by a formula. For example:

$n \equiv 3 \pmod 4$ : with $a = \frac{n+1}{4}$ , we get $\frac4n = \frac1a + \frac{1}{an}$ , and splitting the last term gives three.
$n \equiv 2 \pmod 3$ : with $b = \frac{n+1}{3}$ , we get $\frac4n = \frac1n + \frac1b + \frac{1}{bn}$ .
$n \equiv 5 \pmod 8$ : with $c = \frac{n+3}{8}$ , we get $\frac4n = \frac{1}{2c} + \frac{1}{cn} + \frac{1}{2cn}$ .

Louis Mordell collected identities of this kind for classes modulo 3, 4, 5, 7 and 8. Combined via the Chinese remainder theorem, they settle every $n$ except possibly those with

n \equiv 1,\ 121,\ 169,\ 289,\ 361,\ 529 \pmod{840}.

The smallest prime in these classes is 1009. All six residues are perfect squares mod 840 ( $1^2, 11^2, 13^2, 17^2, 19^2, 23^2$ ), and that is no accident. It is known that polynomial identities of this type can never cover a class of quadratic residues. More congruences shrink the leftover set, but no finite list of identities can close it. A proof needs a different idea.

Exceptions, if any, are rare. In 1970 R. C. Vaughan showed that the number of $n \le N$ without a solution is at most $N \exp\!\left(-c\,(\log N)^{2/3}\right)$ for some constant $c > 0$ . So almost every $n$ works.

How many solutions are there? Write $f(n)$ for the number of solutions. Christian Elsholtz and Terence Tao (2013) proved that for primes $p$ the average of $f(p)$ grows only like a power of $\log p$ . That is plenty to make a counterexample look absurd, but too few for a simple counting argument to rule one out. In 2020 Martin Bright and Daniel Loughran showed that the surfaces $4xyz = n(xy + yz + zx)$ have no Brauer–Manin obstruction, which removes one standard algebraic-geometric reason a solution might fail to exist.

Computer checks. For each $n$ a solution is easy to find, which makes the conjecture easy to test. The search works like this: the smallest denominator satisfies $\frac n4 < x \le \frac{3n}{4}$ . For fixed $x$ , write $\frac4n - \frac1x = \frac ab$ in lowest terms. Then $\frac ab = \frac1y + \frac1z$ exactly when

(ay - b)(az - b) = b^2,

so the pairs $(y, z)$ come from divisors of $b^2$ . The Euclidean algorithm does the reducing. Allan Swett verified the conjecture up to $10^{14}$ in 1999, and Serge E. Salez (2014) pushed the verified range to $n \le 10^{17}$ using modular filters. In August 2025 an arXiv preprint (arXiv:2509.00128) claimed verification up to $10^{18}$ . As far as we can tell it is still a preprint, not a peer-reviewed publication.

Relatives. Wacław Sierpiński (1956) conjectured the same for $\frac5n$ . The broader guess, that for each fixed numerator $k$ the fraction $\frac kn$ is a sum of three unit fractions once $n$ is large enough, is open as well. Like Goldbach and Collatz, the problem is easy to check for any single $n$ but still has no proof for all of them.

Where It Matters

Nobody needs to split 4/n for practical reasons, but the problem works as a small laboratory where many areas of number theory meet:

Diophantine equations. Clearing denominators turns the question into $4xyz = n(xy + yz + zx)$ , a cubic equation that must have positive integer solutions. It is a clean test case for the general question of when such equations have solutions.
Covering congruences. Mordell's approach covers the integers with residue classes that each have their own formula, glued together by the Chinese remainder theorem. The six square classes mod 840 show exactly where this kind of argument stops working.
Analytic number theory. Vaughan's density bound and the Elsholtz–Tao counts use sieve methods and estimates for divisors in residue classes, tools that also appear in work on Goldbach's conjecture.
Algebraic geometry. For each $n$ the equation defines a cubic surface. The Bright–Loughran result places the conjecture inside the modern study of rational and integral points on surfaces.
Algorithms with exact arithmetic. Large-scale verification needs fast factorisation, divisor enumeration and careful sieving by residue class. Egyptian-fraction expansions are also close relatives of continued fractions, another way to break a rational number into simple pieces with integer steps.

Conclusion

The Erdős–Straus conjecture asks a question about fractions that an Egyptian scribe would have understood. Any computer can test it, and for any single $n$ a solution appears almost at once, usually on the first $x$ tried. It has been checked for every $n$ up to $10^{17}$ , and probably $10^{18}$ . It is proved for almost all $n$ in a precise sense and for every $n$ outside six residue classes mod 840. Still, there is no proof that covers all of them.

The obstacle is structural. The identities that handle most numbers cannot handle squares, and counting solutions on average cannot exclude a single exception. Closing the gap will need a new idea. The problem remains open, along with its relatives for $\frac5n$ and beyond.

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The Erdős–Straus Conjecture

Egyptian fractions, six stubborn residue classes, and a 1948 question still open