Durango Bill’s

The Search for a Counterexample

to

Beal’s Conjecture

The search for a
counterexample to Beal's Conjecture. Computer search results.

Does the equation

A

have any solution where:

1) A, B, C, are positive integers

2) x, y, z are positive integers > 2

3) and A, B, C have no common factor?

If you can answer the above question (with either a “Yes” or a “No”), the American Mathematical Society has a $1,000,000 prize waiting for you. http://www.ams.org/profession/prizes-awards/ams-supported/beal-prize

The problem is known as Beal’s Conjecture.

http://www.bealconjecture.com/

http://en.wikipedia.org/wiki/Beal%27s_conjecture

Beal’s Conjecture is that there are no solutions. (A counterexample would be an actual solution to the problem.)

Now for the bad news. If you answer “No”, you have to prove that there are no solutions as per the above. If you answer “Yes”, you have to provide a counterexample.

The “No” answer has so far eluded mathematicians. The “Yes” answer is something that a computer search might find. (At this point, it appears highly unlikely that there are any counterexamples. It also appears highly unlikely that there is any way to prove that the conjecture is true.)

“Solutions” to A

Count A x B y C z gcd(A,B,C)

-------------------------------------------------------

1 2 3 2 3 2 4 2

2 2 4 2 4 2 5 2

3 2 5 2 5 4 3 2

4 2 5 2 5 2 6 2

5 2 6 2 6 2 7 2

6 2 7 2 7 4 4 2

7 2 7 2 7 2 8 2

8 2 8 2 8 8 3 2

9 2 8 2 8 2 9 2

10 4 3 2 6 2 7 2

11 4 3 4 3 2 7 2

12 4 4 2 8 8 3 2

13 4 4 2 8 2 9 2

14 4 4 4 4 8 3 4

15 4 4 4 4 2 9 2

16 6 3 3 3 3 5 3

The table above shows “solutions” up to C

“2” to the “whatever” is a common term in all but one of the above solutions. The equation 2

To win the $1,000,000 prize money, you have to find some combination where the A, B, and C terms don’t have a common factor. (Or alternately, prove that there is no possible answer.) The good news is that this kind of a search is amenable to computer programs.

The Computer
Search

The computer search is relatively easy. First, you compile a table of X

Count X Y(>2) X

--------------------------------

1 1 3 1 Note: “1” to any power = 1

2 2 3 8

3 2 4 16

4 2 5 32

5 2 6 64

6 2 7 128

7 2 8 256

8 2 9 512

9 3 3 27

10 3 4 81

11 3 5 243

12 3 6 729

13 4 3 64

14 4 4 256

15 5 3 125

16 5 4 625

17 6 3 216

18 7 3 343

19 8 3 512

20 9 3 729

21 10 3 1000

To win the $1,000,000, all you have to do is to search the table to find two terms in the X

The author wrote a computer program (actually, a couple of different versions) that can extend the search to larger numbers. Some of the statistics involved are shown in the table below. (Upper Limit = search up to this value for C

Upper Number of Nbr. of A

Limit X

-----------------------------------------------------------

1.0E2 8 36 4

1.0E3 21 231 16

1.0E4 47 1,128 44

1.0E5 93 4,371 73

1.0E6 179 16,110 101

1.0E7 348 60,726 164

1.0E8 679 230,860 228

1.0E9 1,347 907,878 294

1.0E10 2,721 3,703,281 403

1.0E11 5,573 15,531,951 510

1.0E12 11,540 66,591,570 615

1.0E13 24,115 290,778,670 811

1.0E14 50,748 1,287,705,126 1,036

1.0E15 107,363 5,763,460,566 1,320

1.0E16 228,046 26,002,603,081 1,689

1.0E17 485,860 118,030,212,730 2,213

1.0E18 1,037,553 538,258,632,681 2,850

1.0E19 2,219,692 2,463,517,397,278 3,806

1.0E20 4,755,381 11,306,826,605,271 5,200

1.0E21 10,199,002 52,009,825,997,503 7,100

1.0E22 21,893,199 239,656,092,173,400 9,825

1.0E23 47,028,653 1,105,847,125,011,531 13,762

1.0E24 101,078,157 5,108,396,961,797,403 19,581

1.0E25 217,343,187 23,619,030,576,330,078 ?

1.0E26 467,510,251 109,282,917,628,796,626 ?

1.0E27 1,005,918,384 = (N^2 + N)/2 ?

1.0E28 2,164,895,594 N = Nbr. of X^Y terms ?

Computer processing time is roughly proportional to the Nbr. of A

Available computer memory space is also a limiting factor as the search area becomes larger. When the search is extended to C

(There’s an interesting side note that the number of X

Results of the
Computer Search

There are no counterexamples to Beal’s Conjecture with C

The last solution with C

999,999

The last solution where one of the bases (A, B, C) is a prime number is:

310,087

All 19,581 solutions for C

The bad news is that, up through C

For the equation A

4

This can be transformed to:

4

In the above example, the exponent “6” has been transformed to a “3”.

Some examples that disprove this conjecture include:

162

486

Computer hardware for the search consisted of 2 Intel i7, 6-core systems. The computer program language is “C”, and the author’s programs “play with the bits”. One version of the computer algorithm used a 27-bit hash table lookup system. (Another version used remainders from modulo arithmetic.) 11 copies of the program ran in parallel, and each had its own search area. Each program had its own 2GB of RAM. And the current optimized version of the combination still takes 7 weeks running 24/7 to find all combinations up to 10

It appears that no “no common factor” solution exists. However, you have to be a pretty good mathematician to prove it – about $1,000,000 worth of “pretty good”.

Is the $1,000,000
Beal Prize Winnable?

It’s beginning to look like:

1) Finding a counterexample is impossible because there are no counterexamples.

2) Proving that Beal’s Conjecture is true also appears to be impossible.

1) Finding a counterexample may be impossible

If one or more Beal counterexamples existed, they would have to be a subset of the Fermat-Catalan Conjecture.

For additional information, see:

https://en.wikipedia.org/wiki/Fermat%E2%80%93Catalan_conjecture

and

http://mathworld.wolfram.com/Fermat-CatalanConjecture.html

The Fermat-Catalan Conjecture is similar to Beal’s Conjecture except that the following restriction is placed on the exponents.

1/x + 1/y + 1/z < 1.0

The above restriction allows one of the exponents (x, y, z) to be a “2” while Beal’s Conjecture requires all three exponents (x, y, z) to be “3” or greater. The Fermat-Catalan Conjecture implies the number of FCC solutions is finite. The 10 known FCC solutions are shown below.

1) 1

2) 2

3) 7

4) 2

5) 3

6) 17

7) 1414

8) 9262

9) 43

10) 33

The author has confirmed that of the 4,242 solutions to the Fermat Catalan equations up to C

Since Beal Counterexamples (if any exist) are a subset of Fermat Catalan solutions, the number of Beal Counterexamples has to be less than the number of Fermat Catalan solutions.

As the Beal Counterexample search expands, the density of Beal solutions (allowing common factors) becomes increasingly sparse as the number field becomes larger. If one or more Beal Counterexamples existed, something should have shown up somewhere under C

2) Proving the conjecture may be impossible

The limiting factor for the Beal Conjecture is the constraint that a solution must have no common factor among the 3 bases. (The A, B, C). We have seen that there are an infinite number of solutions where common factors exist.

If we look at the equation A

Count P Q A B C

----------------------------------------------------------

1 2 1 3 4 5

2 3 2 5 12 13

3 4 3 7 24 25

4 5 4 9 40 41

5 6 5 11 60 61

6 7 6 13 84 85

N+1 P[N]+1 Q[N]+1 =P

Count P Q A B C

----------------------------------------------------------

1 2 1 3 4 5

2 5 2 21 20 29

3 12 5 119 120 169

4 29 12 697 696 985

5 70 29 4,059 4,060 5,741

6 169 70 23,661 23,660 33,461

N+1 2P[N]+P[N-1] P[N] =P

Count A B C

--------------------------------------------------------

159,154,989 8,005,059 999,967,900 999,999,941

159,154,990 255,405,891 966,833,860 999,999,941

159,154,991 576,010,660 817,442,109 999,999,941

159,154,992 637,064,259 770,810,620 999,999,941

159,154,993 26,497,560 999,648,839 999,999,961

159,154,994 469,112,280 883,138,489 999,999,961

The 3 tables above show Pythagorean right triangles where the bases (A, B, and C) have no common factor. We note that if any pair (AB, AC, or BC) has any common factor, then ordinary algebraic factoring shows that the 3rd element will also share the same common factor.

The first table shows Pythagorean Triangles where the long side and the Hypotenuse differ by just 1. The 2nd table shows solutions where the two sides differ by 1. We also note that any two consecutive integers cannot have a common factor.

The 3rd table shows the last 6 primitive (no common factor) Pythagorean Triples with Side C <= 1,000,000,000. While none of the sides in table 3 are prime numbers, each triple is coprime since there is no common factor in the 3 sides.

The number of coprime (no common factor) Pythagorean solutions expands in direct proportion to the size of the hypotenuse (Side C). With side C (the hypotenuse) <= 1,000,000,000, there are 3,080,075,432 different Pythagorean triangles. Of these, 159,154,994 have coprime sides. See:

https://oeis.org/A101930/internal

and

http://oeis.org/A101931/internal

The above shows that there are no constraints that prevent the bases (A, B, C) from being coprime (no common factor) for the equation: A

Similarly there are no constraints that prevent coprime solutions to Fermat-Catalan equations. However, the larger exponents for the FCC equation (A

To prove that the Beal Conjecture is true, you would have to find some property that forces Beal Counterexamples to be limited to those solutions that have common factors while the Pythagorean and Fermat-Catalan equations are free to include coprime solutions. Without this property, it will be impossible to prove that the Beal Conjecture is true. There doesn’t appear to be any property that forces this result.

Thus it appears that there are no counterexamples to the Beal Conjecture, and at the same time, it may be impossible to prove it. The $1,000,000 Beal prize may be unwinnable.

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