Introduction

Everyone remembers Fermat's Last Theorem: an+bn=cna^{n} + b^{n} = c^{n} has no positive integer solutions once n>2n > 2. It took over 350 years and Andrew Wiles's 1994 proof to close that case. But what happens if you let the three exponents differ?

That is exactly the question behind the Beal conjecture. Consider Ax+By=CzA^{x} + B^{y} = C^{z}, where AA, BB, CC, xx, yy, zz are all positive integers and every exponent is greater than 2. The moment you allow mismatched exponents, solutions actually exist — for example 33+63=353^{3} + 6^{3} = 3^{5}. But look closely: 3, 6 and 3 are not coprime — they all share the factor 3.

The conjecture, first posed by the Texas banker and amateur mathematician Andrew Beal in the 1990s, claims this is never a coincidence: whenever such an equality holds with all exponents above 2, A, B and C must share a common prime factor. Equivalently, if A, B and C are pairwise coprime, no such equality can exist. It sounds like a small tweak on Fermat — but three decades later, nobody has proved it, and nobody has broken it either.

Hunt for Equalities

Below, the computer brute-forces small bases (2 to 20) and exponents (3 to 6) looking for triples where Ax+By=CzA^{x} + B^{y} = C^{z}. Press Search and watch every equality it finds — then check the greatest common factor column.

<p class="hint">Enter positive integer bases A, B and exponents x, y (both above 2) and check whether A^x + B^y lands exactly on some C^z. Or press Search to brute-force many small triples at once.</p>
<div class="inputs">
  <label>A <input id="A" type="number" min="2" max="20" value="3"></label>
  <label>x <input id="X" type="number" min="3" max="8" value="3"></label>
  <label>B <input id="B" type="number" min="2" max="20" value="6"></label>
  <label>y <input id="Y" type="number" min="3" max="8" value="3"></label>
</div>
<div class="btns">
  <button id="check" type="button">Check this pair</button>
  <button id="search" type="button">Search for equalities</button>
  <button id="reset" type="button" class="ghost">Clear</button>
</div>
<div class="status" id="status">Enter A, B, x, y and press Check, or press Search.</div>
<table class="results" id="results">
  <thead>
    <tr><th>Equality found</th><th>gcd(A,B,C)</th></tr>
  </thead>
  <tbody id="resultsBody"></tbody>
</table>

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body { font-family: system-ui, sans-serif; color: #222; margin: 0; }
.hint { font-size: .9rem; color: #444; margin: 0 0 .7rem; line-height: 1.45; }
.inputs { display: flex; gap: .8rem; flex-wrap: wrap; margin-bottom: .6rem; }
.inputs label { font-size: .85rem; font-weight: 600; color: #1d3557; display: flex; flex-direction: column; gap: .2rem; }
.inputs input { width: 64px; font: 600 14px ui-monospace, monospace; padding: .3rem .4rem;
                border: 1px solid #adb1b8; border-radius: 6px; }
.btns { display: flex; gap: .5rem; flex-wrap: wrap; margin-bottom: .6rem; }
button { font: 600 14px system-ui, sans-serif; padding: .45rem .9rem; border: 1px solid #1d3557;
         background: #1d3557; color: #fff; border-radius: 8px; cursor: pointer; }
button.ghost { background: #fff; color: #1d3557; }
.status { font-size: 1rem; font-weight: 600; margin: .4rem 0 .7rem; min-height: 1.4em; }
.status.ok { color: #0a7d33; }
.status.bad { color: #c92f3c; }
table.results { border-collapse: collapse; width: 100%; font: 13px ui-monospace, monospace; }
table.results th, table.results td { border: 1px solid #cdd9e3; padding: .35rem .5rem; text-align: left; }
table.results th { background: #e8eef3; color: #1d3557; }
table.results td.gcd { font-weight: 700; color: #0a7d33; }

const i18nFmt = (s, v) => s.replace(/%(\w+)%/g, (_, k) => String(v[k] ?? ''));
// Beal conjecture: if A^x + B^y = C^z with x, y, z all > 2, the conjecture
// says A, B, C must share a common prime factor. Bounds are kept small so the browser can search instantly.
const MAXBASE = 12, MINEXP = 3, MAXEXP = 5;

function gcd(a, b) { while (b) { [a, b] = [b, a % b]; } return a; }
// Greatest common divisor of three integers, via two pairwise gcd calls.
function gcd3(a, b, c) { return gcd(gcd(a, b), c); }

function setStatus(t, cls) {
  const s = document.getElementById('status');
  s.textContent = t; s.className = 'status ' + cls;
}

function addRow(A, x, B, y, C, z, g) {
  const tb = document.getElementById('resultsBody');
  const tr = document.createElement('tr');
  const eq = document.createElement('td');
  eq.textContent = `${A}^${x} + ${B}^${y} = ${C}^${z}`;
  const gc = document.createElement('td');
  gc.textContent = g;
  gc.className = 'gcd';
  tr.appendChild(eq); tr.appendChild(gc);
  tb.appendChild(tr);
}

// Check whether the single typed-in pair (A^x, B^y) sums to a perfect z-th power for some small z.
function checkOne() {
  const A = parseInt(document.getElementById('A').value, 10);
  const X = parseInt(document.getElementById('X').value, 10);
  const B = parseInt(document.getElementById('B').value, 10);
  const Y = parseInt(document.getElementById('Y').value, 10);
  if (!A || !B || X < 3 || Y < 3) { setStatus('Enter valid integers with exponents of 3 or more.', 'bad'); return; }
  const sum = Math.pow(A, X) + Math.pow(B, Y);
  // Try each exponent z and round-trip check whether the sum is a perfect z-th power.
  for (let z = MINEXP; z <= 8; z++) {
    const C = Math.round(Math.pow(sum, 1 / z));
    for (const cc of [C - 1, C, C + 1]) {
      if (cc >= 2 && Math.pow(cc, z) === sum) {
        const g = gcd3(A, B, cc);
        document.getElementById('resultsBody').innerHTML = '';
        addRow(A, X, B, Y, cc, z, g);
        setStatus(i18nFmt('Found a match — gcd(A,B,C) = %g% (never 1).', { g: g }), 'ok');
        return;
      }
    }
  }
  setStatus('No integer C makes A^x + B^y a perfect power in this range.', 'bad');
}

// Brute-force small A, B in [2,12] and exponents in [3,5], collecting every equality found.
function search() {
  document.getElementById('resultsBody').innerHTML = '';
  setStatus('Searching small triples…', '');
  let checked = 0, found = 0, anyCoprime = false;
  outer:
  for (let A = 2; A <= MAXBASE; A++) {
    for (let x = MINEXP; x <= MAXEXP; x++) {
      const Ax = Math.pow(A, x);
      for (let B = A; B <= MAXBASE; B++) {
        for (let y = MINEXP; y <= MAXEXP; y++) {
          const By = Math.pow(B, y);
          const sum = Ax + By;
          for (let z = MINEXP; z <= MAXEXP; z++) {
            checked++;
            const C = Math.round(Math.pow(sum, 1 / z));
            if (C >= 2 && Math.pow(C, z) === sum) {
              const g = gcd3(A, B, C);
              if (g === 1) anyCoprime = true;
              addRow(A, x, B, y, C, z, g);
              found++;
              if (found >= 12) break outer;
            }
          }
        }
      }
    }
  }
  if (anyCoprime) setStatus('Coprime counterexample found! (This would disprove the conjecture — check your numbers.)', 'bad');
  else setStatus(i18nFmt('Found %n% equalities after checking %k% combinations — every single one shares a common factor.', { n: found, k: checked }), 'ok');
}

document.getElementById('check').addEventListener('click', checkOne);
document.getElementById('search').addEventListener('click', search);
document.getElementById('reset').addEventListener('click', () => {
  document.getElementById('resultsBody').innerHTML = '';
  setStatus('Enter A, B, x, y and press Check, or press Search.', '');
});
Result
Console

Notice what never happens: no hit in the table is ever pairwise coprime. Every single equality the brute force turns up — like 33+63=353^{3} + 6^{3} = 3^{5} or 63+123=64⋅36^{3} + 12^{3} = 6^{4}\cdot 3-style scalings — shares a prime among A, B and C. Checking one candidate triple is instant arithmetic. Confirming that a coprime counterexample can never exist, for every triple of exponents above 2, is the open question — and no brute force, however large, can ever finish that search.

The Real Difficulty

How hard is the Beal conjecture, really? Not the searching — the proving.

  • Checking a candidate triple is trivial: raise A to x, B to y, add them, compare to C raised to z, then take the greatest common factor of A, B and C. A computer does this instantly, and exhaustive searches into the trillions have never found a counterexample.
  • The conjecture itself is OPEN. Posed by Andrew Beal around 1993, it asks whether Ax+By=CzA^{x} + B^{y} = C^{z} with x,y,z>2x, y, z > 2 ever has a solution where A, B, C are pairwise coprime. Nobody has proved it can't happen, and nobody has found a single triple where it does.
  • The million-dollar incentive. Beal, a banker and amateur number theorist, personally funds a prize — now US $1,000,000 — held in trust by the American Mathematical Society (AMS), awarded for a peer-reviewed, published proof or a published counterexample, after a two-year period during which the mathematical community can scrutinize the claim.
  • No shortcut exists. Like Fermat's Last Theorem before it, Beal's claim is a statement about infinitely many possible triples — no amount of brute-force checking of finite ranges can ever certify that no counterexample exists anywhere in that infinite space.

This is the same kind of hardness that stalls the abc conjecture: the obstacle isn't computing anything, it's proving a universal claim about the integers — a very different mountain from the ones climbed by algorithms in P vs NP.

Where It Matters

The Beal conjecture sits at the center of a family of hard questions about sums of powers:

  • A generalization of Fermat's Last Theorem: setting x=y=zx = y = z recovers exactly Fermat's equation, so Beal's conjecture — if proved — would explain why Fermat's theorem holds as a special case of a much broader pattern.
  • Diophantine equations: Beal is a pure statement about integer solutions to a polynomial-like equation, the same family of questions explored in Diophantine equations, though Beal is conjectured rather than proved undecidable.
  • The role of common factors: the conjecture highlights how deeply prime factorization constrains which sums of perfect powers can ever line up — a theme that echoes the abc conjecture's radical/quality framework.
  • A magnet for amateur and professional mathematicians alike: the prize, the simple statement and the total absence of a proof make Beal one of the most approachable "big" open problems to explain — and one of the hardest to actually solve.

A proof would be a landmark in number theory; a counterexample would instantly reshape how mathematicians think about powers, sums and shared factors.

Conclusion

The Beal conjecture takes Fermat's famous equation, lets the exponents differ, and asks a single clean question: must A, B and C always share a factor once x,y,z>2x, y, z > 2? Every computer search ever run says yes. No one has shown it must be true for all infinitely many possible triples.

So the equation sits there, deceptively small, with US $1,000,000 waiting in trust at the American Mathematical Society for whoever settles it — proof or counterexample. Like P vs NP and the abc conjecture, Beal is a reminder that some of the most expensive questions in mathematics can be written in a single line, and checked in an instant — while proving them forever might take another century.

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