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The Search for a Counterexample
to
Beal’s Conjecture


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


Does the equation
Ax + By = Cz
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 Ax + By = Cz with Cz <= 1000 (but A, B, C have a common factor)

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 Cz <= 1,000. However, we note that for each of the above 16 “solutions”, the “A”, “B”, and “C” integers can all be divided by the gcd (Greatest Common Divisor).

   “2” to the “whatever” is a common term in all but one of the above solutions. The equation 2N + 2N = 2N+1 can be used to form an infinite number of “solutions”. (Simply use any N > 2.) However, all of these have a common factor of 2 (or some multiple of 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 XY terms. Then you just search the results to see if the sum of any 2 terms matches something else. The table below shows all possible XY combinations up to 1,000.

Count       X       Y(>2)     XY
--------------------------------
  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 XY column such that their sum is equal to some other term. For example, if we use row 17 (X = 6) and row 9 (X = 3), the XY sum for these two terms is 216 + 27 = 243. The 243 matches the 243 in row 11. The bad news is that the three Xs involved (6, 3, and 3) all have a common factor such that they can be divided by 3.

   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 Cz in Ax + By = Cz.)

Upper       Number of          Nbr. of Ax + By     Nbr. of
Limit        XY Terms            Candidates       Solutions
-----------------------------------------------------------
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 Ax + By Candidates.  Thus the run time for each new row is about 4 1/2 to 5 times longer than the row above it. (The actual ratio appears to approach 10^(2/3) ~= 4.64. It’s actually worse for the largest powers due to increased hash table densities.)

   Available computer memory space is also a limiting factor as the search area becomes larger. When the search is extended to CZ <= 1.0E24, there are 101,078,157 rows in the XY table. Relevant information has to be stored in the computer for all 101,078,157 of these. If you tried to physically print the XY list using single spacing typing at 6 lines per inch, the printed list would be over 265 miles long.

(There’s an interesting side note that the number of XY terms that are less than the limit seems to asymptotically approach the cube root of the limit.)



Results of the Computer Search

   There are no counterexamples to Beal’s Conjecture with CZ <= 1.0E24. A variation of the program assumed that at least one of the exponents ("X" and "Y") must be >= 5. this allowed a search to 1.0E27. There were no counterexamples within this search range.

The last solution with CZ under 1.0E24 is:
999,9994 + 999,9993 = 99,999,9003  (= 9.99997E23)  gcd(A, B, C) = 999,999

The last solution where one of the bases (A, B, C) is a prime number is:
310,0874 + 99,537,9273 = 99,848,0143 (=9.95447E23)  gcd(A, B, C) = 310,087   (310,087 is prime)

  All 19,581 solutions for Cz <= 1.0E24 can be seen in the file http://durangobill.com/Beals24simple.txt . Data in the file has been limited to just the Count and the variables A, x, B, y, C, z so that it can be easily read into an Excel file. You will probably want to add additional columns for Ax, By, Cz, gcd(A, B, C), etc.

   The bad news is that, up through Cz <= 1,000,000,000,000,000,000,000,000, each of the 19,581 “Solutions” has some common factor. . To win the $1,000,000 prize (for a counterexample), some solution must be found where the A, B, and C terms don’t have a common factor.

   For the equation AX + BY = CZ, there has been some speculation that two of the exponents (X,Y,Z) must be the same (as per above) – or that they can be transformed into an equivalent expression where they are the same. As an example of something that can be transformed we have:
43 + 26 = 27
This can be transformed to:
43 + (22)3 = 27
In the above example, the exponent “6” has been transformed to a “3”.

 Some examples that disprove this conjecture include:
1623 + 274 = 97
4863 + 275 = 317

   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 1024.

   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)  1m + 23 = 32      (m > 6)
2)  25 + 72 = 34
3)  73 + 132 = 29
4)  27 + 173 = 712
5)  35 + 114 = 1222

6)  177 + 762713 = 210639282
7)  14143 + 22134592 = 657
8)  92623 + 153122832 = 1137
9)  438 + 962223 = 300429072
10)  338 + 15490342 = 156133

   The author has confirmed that of the 4,242 solutions to the Fermat Catalan equations up to CZ <= 1.0E16, the 10 solutions shown above are the only 10 where A, B, and C are coprime.

   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 CZ <= 1.0E24. It appears that the subset size of Beal Counterexamples is “0”.


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 A2 + B2 = C2 (The 2 sides and hypotenuse of a Pythagorean right triangle), there are an infinite number of solutions where A, B, and C do not have a common factor. For example:

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      =P2-Q2      =2PQ      =P2+Q2

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]     =P2-Q2      =2PQ      =P2+Q2

   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: A2 + B2 = C2.

   Similarly there are no constraints that prevent coprime solutions to Fermat-Catalan equations. However, the larger exponents for the FCC equation (Ax + By = Cz) reduce the number of solutions to a small finite number.

   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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