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
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 unless A,B,C have a common divisor > 1.
Now for the bad news. If you answer “No”, you have to prove that there are no counterexamples 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 23,619,030,793,673,266 ?
1.0E26 467,510,251 109,282,918,096,306,878 ?
1.0E27 1,005,918,386 505,935,900,149,381,691 ?
1.0E28 2,164,895,594 2,343,386,467,542,754,215 ?
1.0E29 4,660,092,821 = (N^2 - N)/2 + N
1.0E30 10,032,752,675 N = Nbr. of X^Y terms
In the above table “Nbr. of Ax + By Candidates” is the number of combinations that a computer program has to test to see if the result produces a valid solution for the equation Ax + By = Cz. The calculation is equal to COMBIN( Number of Xy, 2) terms + N. (The “+N” factor allows Ax to equal By - for example 23 + 23 = 24).
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.) (The current algorithm for the Threadripper search cuts this ratio to about 3.83 per row.)
Available computer memory space is also a limiting factor as the search area becomes larger. When the search is extended to CZ <= 1.0E30, there are 10,032,752,675 rows in the XY table. Relevant information has to be stored in the computer for all 10,032,752,675 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 26,390 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 <= 1.0E28. A variation of the program assumed that at least one of the exponents ("X" and/or "Y") must be >= 5. This allowed a search to 1.0E30. There are no counterexamples with CZ <= 1.0E30 (Assumes at least one of the exponents ("X", "Y") is > 4). (This program variation required 10,032,752,675 rows in XY table - and about 108 GB of total RAM were used for some very large arrays. The program ran for over 2 months on an AMD Threadripper 2920X processor and used all 12 cores via multi-threading.)
No solution exists if the exponents are (3,3,3) or (4,4,4) because of "Fermat's Last Theorem". Solutions exist if the exponents are (3,3,4) or (3,4,4) in any order, but the Wikipedia article on Beal's Conjecture ( https://en.wikipedia.org/wiki/Beal_conjecture ) references sources that prove that these solutions can not be counterexamples. Thus, at least one of the exponents ("X", "Y", "Z") must contain a "5" or higher.
If a counterexample exists, the exponents "X" and/or "Y" must contain at least one exponent >= "4". A search out to Cz < 1.0E28 was made with the restriction that exponent "X" was >= 4.
51,660 general solutions to Ax + By = Cz were found. All of these had a GCD(A, B, C) >= 2 - hence none of them is a counterexample. The complete list can be seen at: http://www.durangobill.com/BealXgt3e28.pdf (Large file 5.4 MB)
The last 10 (sorted by Cz) of these general solutions are:
Count Base Power Base Power Base Power CZ GCD
A X B Y C Z (A, B, C)
51,651 9,985,740 4 356,158,060 3 2,153,591,260 3 9.9883E+27 3,328,580
51,652 8,251,453 4 1,749,308,036 3 2,153,629,233 3 9.9888E+27 8,251,453
51,653 9,363,779 4 1,320,292,839 3 2,153,669,170 3 9.9893E+27 9,363,779
51,654 9,446,248 4 1,265,797,232 3 2,153,744,544 3 9.9904E+27 9,446,248
51,655 6,194,994 4 2,042,283,022 3 2,153,792,914 3 9.9911E+27 2,064,998
51,656 9,994,896 4 234,880,056 3 2,153,900,088 3 9.9926E+27 4,997,448
51,657 7,720,118 4 1,860,548,438 3 2,153,912,922 3 9.9927E+27 7,720,118
51,658 9,790,625 4 930,109,375 3 2,153,937,500 3 9.9931E+27 9,790,625
51,659 9,574,777 4 1,168,122,794 3 2,154,324,825 3 9.9985E+27 9,574,777
51,660 9,973,873 4 468,772,031 3 2,154,356,568 3 9.9989E+27 9,973,873
Another variation of the program assumed that at least one of the exponents ("X", "Y") must be >= 5. A computer search out to Cz <= 1.0E30 was made. 6,425 general solutions to Ax + By = Cz were found, but all of these had a GCD(A,B,C) >= 2, and are thus not counterexamples. The 6,425 solutions can be seen at http://www.durangobill.com/BealXgt4e30.pdf . Here are the last 10 of these:
Count Base Power Base Power Base Power CZ GCD
A X B Y C Z (A, B, C)
6416 999 9 998,001 5 9,970,029,990 3 9.9104E+29 999
6417 999 10 997,002,999 3 9,970,029,990 3 9.9104E+29 999
6418 998,001 5 997,002,999 3 9,970,029,990 3 9.9104E+29 998,001
6419 971,074 5 18,935,943 4 31,559,905 4 9.9207E+29 485,537
6420 18,557 7 6,198,520,482 3 9,986,505,221 3 9.9596E+29 18,557
6421 69,312 6 9,608,306,688 3 31,606,272 4 9.9791E+29 69,312
6422 99,968 6 199,936 5 999,680 5 9.9840E+29 99,968
6423 199,936 5 9,993,601,024 3 999,680 5 9.9840E+29 199,936
6424 99,999 6 99,999 5 999,990 5 9.9995E+29 99,999
6425 99,999 5 9,999,800,001 3 999,990 5 9.9995E+29 99,999
For the equation AX + BY = CZ, there has been some speculation that two of the exponents (X,Y,Z) must be the same – 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
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 9,725 solutions to the Fermat Catalan Conjecture up to CZ <= 1.0E19, the 10 solutions shown above are the only 10 where A, B, and C are coprime. The 9,725 solutions can be seen here. http://www.durangobill.com/FermatCatalan.pdf
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 <= 1.0E28. 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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