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Famous results and impossibility proofs

This page is the showcase. The additive equations below — sums of like powers, taxicab numbers, the Euler brick — are the problems whose individual solutions made history, and several were first cracked by famous computer searches. Mathilda solves each with a purpose-built method (meet-in-the-middle, modular descent, divisor enumeration), and closes the page with equations whose answer is the empty set: a proof that no solution exists at all.

In[1]:= Solve[x^5 + y^5 + z^5 + w^5 == r^5 && 0 < x < y < z < w < r < 700, {x, y, z, w, r}, Integers]
Out[1]= {{x -> 27, y -> 84, z -> 110, w -> 133, r -> 144}, {x -> 54, y -> 168, z -> 220, w -> 266, r -> 288}, {x -> 81, y -> 252, z -> 330, w -> 399, r -> 432}, {x -> 108, y -> 336, z -> 440, w -> 532, r -> 576}}

1. Lander and Parkin — Euler's conjecture falls

In 1769 Euler conjectured that a \(k\)-th power needs at least \(k\) other \(k\)-th powers to sum to it — no three fourth powers add to a fourth power, no four fifth powers to a fifth power. In 1966, using a CDC 6600, Lander and Parkin found the counterexample that ended the fifth-power case: [ 27^5 + 84^5 + 110^5 + 133^5 = 144^5. ] That is the first tuple in In[1] above; the other three are its \(2\times\), \(3\times\), and \(4\times\) multiples inside the search box. Mathilda finds them with an ordering-aware meet-in-the-middle search: it tabulates the partial sums of one half of the variables and binary-searches for the complement, carrying 128-bit integers so the fifth powers do not overflow. Tightening the box to isolate the primitive solution returns it alone:

In[1]:= Solve[x^5 + y^5 + z^5 + w^5 == r^5 && 0 < x <= y <= z <= w < 150 && r < 200, {x, y, z, w, r}, Integers]
Out[1]= {{x -> 27, y -> 84, z -> 110, w -> 133, r -> 144}}

2. Frye's fourth-power counterexample

The fourth-power case of Euler's conjecture held out until 1988, when Roger Frye, using Elkies' elliptic-curve analysis to localise the search, found the smallest solution on a Connection Machine: [ 95800^4 + 217519^4 + 414560^4 = 422481^4. ] Mathilda reaches it directly via Frye's own mod-5 descent — the divisibility \(625 \mid (w^4 - C^4)\) plus secondary moduli prune the \((w, C)\) pairs — over a narrow window around the known \(w\):

In[1]:= Solve[x^4 + y^4 + z^4 == w^4 && 0 < x < y < z < w && 422400 < w < 422500, {x, y, z, w}, Integers]
Out[1]= {{x -> 95800, y -> 217519, z -> 414560, w -> 422481}}

Search windows

The narrow window keeps this example under a second. The full cold search over \(0 < w < 10^6\) is a single-core computation of roughly ten minutes — Frye's original run used a 65 536-processor machine. The performance page puts the timing in context.

3. Taxicab numbers — two cubes, two ways

Ramanujan's taxicab number \(1729\) is the smallest integer expressible as a sum of two cubes in two different ways: \(1729 = 1^3 + 12^3 = 9^3 + 10^3\). For a fixed target, Mathilda uses the divisor method — for a sum of two cubes, \(x + y\) divides the target, so the answer comes from the divisors of \(1729\), not a two-dimensional scan:

In[1]:= Solve[x^3 + y^3 == 1729 && 0 < x <= y && x < 10^5, {x, y}, Integers]
Out[1]= {{x -> 1, y -> 12}, {x -> 9, y -> 10}}

Asked directly for two distinct representations, the same number falls out of a meet-in-the-middle search:

In[2]:= Solve[x^3 + y^3 == z^3 + w^3 && 0 < x < y && 0 < z < w && x != z && x^3 + y^3 < 1730, {x, y, z, w}, Integers]
Out[2]= {{x -> 1, y -> 12, z -> 9, w -> 10}, {x -> 9, y -> 10, z -> 1, w -> 12}}

4. The Euler brick

An Euler brick is a rectangular box whose three edges and all three face diagonals are integers: \(x^2 + y^2 = a^2\), \(x^2 + z^2 = b^2\), \(y^2 + z^2 = c^2\). Mathilda solves the coupled system by staged elimination — only the free edge variables are enumerated, and each diagonal is closed as an exact square root per candidate:

In[1]:= Solve[x^2 + y^2 == a^2 && x^2 + z^2 == b^2 && y^2 + z^2 == c^2 && 0 < x < y < z < 250 && a > 0 && b > 0 && c > 0, {x, y, z, a, b, c}, Integers]
Out[1]= {{x -> 44, y -> 117, z -> 240, a -> 125, b -> 244, c -> 267}}

The smallest Euler brick, discovered by Paul Halcke in 1719, has edges \((44, 117, 240)\).

5. Two more classics

The Markov equation \(x^2 + y^2 + z^2 = 3xyz\) generates the Markov triples, and Mathilda counts thirteen with all entries \(\le 1000\); Brocard's problem \(n! + 1 = m^2\) has only the three known Brown numbers:

In[1]:= Length[Solve[x^2 + y^2 + z^2 == 3 x y z && 0 < x <= y <= z <= 1000, {x, y, z}, Integers]]
Out[1]= 13

In[2]:= Solve[Factorial[n] + 1 == m^2 && n > 0 && m > 0 && n < 100, {n, m}, Integers]
Out[2]= {{n -> 4, m -> 5}, {n -> 5, m -> 11}, {n -> 7, m -> 71}}

\(4! + 1 = 25\), \(5! + 1 = 121\), \(7! + 1 = 5041 = 71^2\) — whether any fourth Brown number exists is still open, so the bounded query is the honest statement: these are all with \(n < 100\).

6. When the answer is a proof of impossibility

The deepest results are the ones where the solution set is empty, and Mathilda returns {} only when it can prove it. The supreme example is Fermat's Last Theorem — that \(x^n + y^n = z^n\) has no positive-integer solution for \(n \ge 3\), proved by Andrew Wiles in 1995. Mathilda recognises the form and returns the proof instantly:

In[1]:= Solve[x^3 + y^3 == z^3 && x > 0 && y > 0 && z > 0, {x, y, z}, Integers]
Out[1]= {}

That single {} stands for one of the most celebrated theorems in mathematics. The Erdős–Straus / Egyptian-fraction equation, by contrast, is solvable, and Mathilda's reciprocal recursion returns every ordered unit-fraction decomposition of \(1\) into three parts:

In[2]:= Solve[1 == 1/x + 1/y + 1/z && 0 < x <= y <= z, {x, y, z}, Integers]
Out[2]= {{x -> 2, y -> 3, z -> 6}, {x -> 2, y -> 4, z -> 4}, {x -> 3, y -> 3, z -> 3}}

The three ways to write \(1\) as a sum of three unit fractions: \(\tfrac12 + \tfrac13 + \tfrac16\), \(\tfrac12 + \tfrac14 + \tfrac14\), and \(\tfrac13 + \tfrac13 + \tfrac13\).

References

  1. L. J. Lander & T. R. Parkin, Counterexample to Euler's conjecture on sums of like powers, Bull. Amer. Math. Soc. 72 (1966), 1079.
  2. R. Frye, Finding \(95800^4 + 217519^4 + 414560^4 = 422481^4\) on the Connection Machine, Proc. Supercomputing '88, IEEE, 1988, 106–116.
  3. N. D. Elkies, On \(A^4 + B^4 + C^4 = D^4\), Math. Comp. 51 (1988), 825–835.
  4. A. Wiles, Modular elliptic curves and Fermat's Last Theorem, Ann. of Math. 141 (1995), 443–551.

Next: How Mathilda compares — the head-to-head against sympy and PARI/GP, with the coverage table and the honest gaps.