# # * It is valid to use proof by contradiction: assume the conjecture is true, and show that that leads to a contradiction. It is not valid to use proof by circular reasoning: assume the conjecture is true, put in some irrelevant steps, and show that it follows that the conjecture is true. # # #
#
# * A valid counterexample needs to satisfy all four conditions—don't leave one out:
#
# > $A, B, C, x, y, z$ are positive integers
# $x, y, z > 2$
# $A^x + B^y = C^z$
# $A, B, C$ have no common prime factor.
#
# (If you think you have a valid counterexample, before you share it with Andrew Beal, or me, or anyone else, you can check it with my [Online Beal Counterexample Checker](http://norvig.com/bealcheck.html).)
#
#
# # * One correspondent claimed that $27^4 + 162 ^ 3 = 9 ^ 7$ was a solution, because the first three conditions hold, and the common factor is 9, which isn't a prime. But of course, if $A, B, C$ have 9 as a common factor, then they also have 3, and 3 is prime. The phrase "no common prime factor" means the same thing as "no common factor greater than 1." # #
# # * Another claimed that $2^3+2^3=2^4$ was a counterexample, because all the bases are 2, which is prime, and prime numbers have no prime factors. But that's not true; a prime number has itself as a factor. # #
# # * A creative person offered $1359072^4 - 940896^4 = 137998080^3$, which fails both because $3^3 2^5 11^2$ is a common factor, and because it has a subtraction rather than an addition (although, as Julius Jacobsen pointed out, that can # #
# # * Another Beal fan started by saying "Let $C = 43$ and $z = 3$. Since $43 = 21 + 22$, we have $43^3 = (21^3 + 22^3).$" But of course $(a + b)^3 \ne (a^3 + b^3)$. This fallacy is called [the freshman's dream](https://en.wikipedia.org/wiki/Freshman%27s_dream) (although I remember having different dreams as a freshman). # #
# # * Multiple people proposed answers similar to this one: # In[1]: from math import gcd #### In Python versions < 3.5, use "from fractions import gcd" # In[2]: A, B, C = 60000000000000000000, 70000000000000000000, 82376613842809255677 x = y = z = 3. A ** x + B ** y == C ** z and gcd(gcd(A, B), C) == 1 # **WOW!** The result is `True`! Is this a real counterexample to Beal? And also a disproof of Fermat? # # Alas, it is not. Notice the decimal point in "`3.`", indicating a floating point number, with inexact, limited precision. Change the inexact "`3.`" to an exact "`3`" and the result changes to "`False`". Below we see that the two sides of the equation are the same for the first 18 digits, but differ starting with the 19th: # In[3]: (A ** 3 + B ** 3, C ** 3) # They say "close" only counts in horseshoes and hand grenades, and if you threw two horseshoes at a stake on the planet [Kapteyn-b](https://en.wikipedia.org/wiki/Kapteyn_b) (a possibly habitable and thus possibly horseshoe-playing exoplanet 12.8 light years from Earth) and the two paths differed in the 19th digit, the horseshoes would end up [less than an inch](https://www.google.com/webhp?sourceid=chrome-instant&ion=1&espv=2&ie=UTF-8#q=12.8%20light%20years%20*%201e-19%20in%20inches) apart. That's really, really close, but close doesn't count in number theory. # # # *The Simpsons* and Fermat # # Speaking of *close*: in two different [episodes of *The Simpsons*](http://www.npr.org/sections/krulwich/2014/05/08/310818693/did-homer-simpson-actually-solve-fermat-s-last-theorem-take-a-look), close counterexamples to Fermat's Last Theorem are shown: # $1782^{12} + 1841^{12} = 1922^{12}$ and $3987^{12} + 4365^{12} = 4472^{12}$. These were designed by *Simpsons* writer David X. Cohen to be correct up to the precision found in most handheld calculators. Cohen found the equations with a program that must have been something like this: # In[4]: from itertools import combinations def simpsons(bases, powers): """Find the integers (A, B, C, n) that come closest to solving Fermat's equation, A ** n + B ** n == C ** n. Let A, B range over all pairs of bases and n over all powers.""" equations = ((A, B, iroot(A ** n + B ** n, n), n) for A, B in combinations(bases, 2) for n in powers) return min(equations, key=relative_error) def iroot(i, n): "The integer closest to the nth root of i." return int(round(i ** (1./n))) def relative_error(equation): "Error between LHS and RHS of equation, relative to RHS." (A, B, C, n) = equation LHS = A ** n + B ** n RHS = C ** n return abs(LHS - RHS) / RHS # In[5]: simpsons(range(1000, 2000), [11, 12, 13]) # In[6]: simpsons(range(3000, 5000), [12]) # # Back to Beal: `beal` 2.0 and 2.1 # # In October 2015 I looked back at my original [program from 2000](http://norvig.com/beal2000.html). # I ported it from Python 1.5 to 3.5 (by putting parens around the argument to `print` and adding `long = int`). It runs 250 times faster today, a tribute to both computer hardware engineers and the developers of the Python interpreter. # # I found that I had misunderstood [the problem in 2000](https://web.archive.org/web/19991127081319/http://bealconjecture.com/). I thought that, *by definition*, $A$ and $B$ could not have a common factor, but actually, the definition of the conjecture only rules out examples where all three of $A, B, C$ share a common factor. I rewrote the program to reflect that, but then [Mark Tiefenbruck ](mailto:mark @tiefenbruck.org) (and later Edward P. Berlin and Shen Lixing) wrote to point out that my original program was actually correct, not by definition, but by derivation: if $A$ and $B$ have a commmon prime factor $p$, then the sum of $A^x + B^y$ must also have that factor $p$, and since $A^x + B^y = C^z$, then $C^z$ and hence $C$ must have the factor $p$. So I was wrong twice—I originally failed to understand the problem completely, and then I failed to recognize the optimization—and that means the original program was correct. # # Mark Tiefenbruck also suggested another optimization: only consider exponents that are odd primes, or 4. The idea is that a number like 512 can be expressed as either $2^9$ or $8^3$, and my program doesn't need to consider both. In general, any time we have a composite exponent, such as $b^{qp}$, where $p$ is prime, we should ignore $A=b, x=qp$, and instead consider only $A=b^q, x=p$. There's one complication to this scheme: 2 is a prime, but 2 is not a valid exponent for a Beal counterexample. So we will allow 4 as an exponent, as well as all odd primes up to `max_x`. # # Here is the complete, updated program: # In[7]: from math import gcd, log from itertools import combinations, product def beal(max_A, max_x): """See if any A ** x + B ** y equals some C ** z, with gcd(A, B) == 1. Consider any 1 <= A,B <= max_A and x,y <= max_x, with x,y prime or 4.""" Apowers = make_Apowers(max_A, max_x) Czroots = make_Czroots(Apowers) for (A, B) in combinations(Apowers, 2): if gcd(A, B) == 1: for (Ax, By) in product(Apowers[A], Apowers[B]): Cz = Ax + By if Cz in Czroots: C = Czroots[Cz] x, y, z = exponent(Ax, A), exponent(By, B), exponent(Cz, C) print('{} ** {} + {} ** {} == {} ** {} == {}' .format(A, x, B, y, C, z, C ** z)) def make_Apowers(max_A, max_x): "A dict of {A: [A**3, A**4, ...], ...}." exponents = exponents_upto(max_x) return {A: [A ** x for x in (exponents if (A != 1) else [3])] for A in range(1, max_A+1)} def make_Czroots(Apowers): return {Cz: C for C in Apowers for Cz in Apowers[C]} def exponents_upto(max_x): "Return all odd primes up to max_x, as well as 4." exponents = [3, 4] if max_x >= 4 else [3] if max_x == 3 else [] for x in range(5, max_x, 2): if not any(x % p == 0 for p in exponents): exponents.append(x) return exponents def exponent(Cz, C): """Recover z such that C ** z == Cz (or equivalently z = log Cz base C). For exponent(1, 1), arbitrarily choose to return 3.""" return 3 if (Cz == C == 1) else int(round(log(Cz, C))) # It takes less than a second to verify that there are no counterexamples for combinations up to $100^{100}$, a computation that took Andrew Beal thousands of hours on his 1990s-era computers: # In[8]: get_ipython().run_line_magic('time', 'beal(100, 100)') # The execution time goes up roughly with the square of `max_A`, so with 5 times more `A` values, this computation takes about 25 times longer: # In[9]: get_ipython().run_line_magic('time', 'beal(500, 100)') # # How `beal` Works # # The function `beal` first does some precomputation, creating two data structures: # * `Apowers`: a dict of the form `{A: [A**3, A**4, ...]}` giving the # nonredundant powers (prime and 4th powers) of each base, `A`, from 1 to `max_x`. # * `Czroots`: a dict of `{C**z : C}` pairs, giving the zth root of each power in `Apowers`. # # Then we consider all combinations of two bases, `A` and `B`, from `Apowers`. # Here is a very small example Apowers table: # In[10]: Apowers = make_Apowers(6, 10) Apowers # Consider the combination where `A` is `3` and `B` is `6`. Of course `gcd(3, 6) == 3`, so the program would not consider them further, but imagine if they did not share a common factor. Then we would look at all possible `Ax + By` sums, for `Ax` in `[27, 81, 243, 2187]` and `By` in `[216, 1296, 7776, 279936].` One of these would be `27 + 216`, which sums to `243`. We look up `243` in `Czroots`: # In[11]: Czroots = make_Czroots(Apowers) print(Czroots) Czroots[243] # We see that `243` is in `Czroots`, with value `3`, so this would be a counterexample (except for the common factor). The program uses the `exponent` function to recover the values of `x, y, z`, and prints the results. # # # Is the Program Correct? # # Can we gain confidence in the program? It is difficult to test `beal`, because the expected output is nothing, for all known inputs. # One thing we can do is verify that `beal` finds cases like `3 ** 3 + 6 ** 3 == 3 ** 5 == 243` that would be a counterexample except for the common factor `3`. We can test this by temporarily replacing the `gcd` function with a mock function that always reports no common factors: # In[12]: def gcd(a, b): return 1 beal(100, 100) # Let's make sure all those expressions are true: # In[13]: {3 ** 3 + 6 ** 3 == 3 ** 5 == 243, 7 ** 7 + 49 ** 3 == 98 ** 3 == 941192, 8 ** 4 + 16 ** 3 == 2 ** 13 == 8192, 8 ** 5 + 32 ** 3 == 16 ** 4 == 65536, 9 ** 3 + 18 ** 3 == 9 ** 4 == 6561, 16 ** 5 + 32 ** 4 == 8 ** 7 == 2097152, 17 ** 4 + 34 ** 4 == 17 ** 5 == 1419857, 19 ** 4 + 38 ** 3 == 57 ** 3 == 185193, 27 ** 3 + 54 ** 3 == 3 ** 11 == 177147, 28 ** 3 + 84 ** 3 == 28 ** 4 == 614656, 34 ** 5 + 51 ** 4 == 85 ** 4 == 52200625} # I get nervous having an incorrect version of `gcd` around; let's change it back, quick! # In[14]: from math import gcd beal(100, 100) # We can also provide some test cases for the subfunctions of `beal`: # In[15]: def tests(): assert make_Apowers(6, 10) == { 1: [1], 2: [8, 16, 32, 128], 3: [27, 81, 243, 2187], 4: [64, 256, 1024, 16384], 5: [125, 625, 3125, 78125], 6: [216, 1296, 7776, 279936]} assert make_Czroots(make_Apowers(5, 8)) == { 1: 1, 8: 2, 16: 2, 27: 3, 32: 2, 64: 4, 81: 3, 125: 5, 128: 2, 243: 3, 256: 4, 625: 5, 1024: 4, 2187: 3, 3125: 5, 16384: 4, 78125: 5} Czroots = make_Czroots(make_Apowers(100, 100)) assert 3 ** 3 + 6 ** 3 in Czroots assert 99 ** 97 in Czroots assert 101 ** 100 not in Czroots assert Czroots[99 ** 97] == 99 assert exponent(10 ** 5, 10) == 5 assert exponent(7 ** 3, 7) == 3 assert exponent(1234 ** 999, 1234) == 999 assert exponent(12345 ** 6789, 12345) == 6789 assert exponent(3 ** 10000, 3) == 10000 assert exponent(1, 1) == 3 assert exponents_upto(2) == [] assert exponents_upto(3) == [3] assert exponents_upto(4) == [3, 4] assert exponents_upto(40) == [3, 4, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37] assert exponents_upto(100) == [ 3, 4, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97] assert gcd(3, 6) == 3 assert gcd(3, 7) == 1 assert gcd(861591083269373931, 94815872265407) == 97 assert gcd(2*3*5*(7**10)*(11**12), 3*(7**5)*(11**13)*17) == 3*(7**5)*(11**12) return 'tests pass' tests() # The program is mostly straightforward, but relies on the correctness of these arguments: # # * Are we justified in taking `combinations` without replacements from the table? In other words, are we sure there are no solutions of the form $A^x + A^x = C^z$? Yes, we can be sure, because then $2\;A^x = C^z$, and all the factors of $A$ would also be factors of $C$. # #
# * Are we justified in having a single value for each key in the `Czroots` table? Consider that $81 = 3^4 = 9^2$. We put `{81: 3}` in the table and discard `{81: 9}`, because any number that has 9 as a factor will always have 3 as a factor as well, so 3 is all we need to know. But what if a number could be formed with two bases where neither was a multiple of the other? For example, what if $2^7 = 5^3 = s$; then wouldn't we have to have both 2 and 5 as values for $s$ in the table? Fortunately, that can never happen, because of the [fundamental theorem of arithmetic](https://en.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic). # #
# * Could there be a rounding error involving the `exponent` function that was not caught by the tests? Possibly; but `exponent` is not used to find counterexamples, only to print them, so any such error wouldn't cause us to miss a counterexample. # #
# * Are we justified in only considering exponents that are odd primes, or the number 4? In one sense, yes, because when we consider the two terms $A^{qp}$ and $(A^q)^p$, we find they are always equal, and always have the same prime factors (the factors of $A$), so for the purposes of the Beal problem, they are equivalent, and we only need consider one of them. In another sense, there is a difference. With this optimization, when we run `beal(6, 10)`, we are no longer testing $512$ as a value of $A$ or $B$, even though $512 = 2^9$ and both $2$ and $9$ are within range, because the program chooses to express $512$ as $8^3$, and $8$ is not in the specified range. So the program is still correctly searching for counterexamples, but the space that it searches for given `max_A` and `max_x` is different with this optimization. # #
# * Are we really sure that when $A$ and $B$ have a common factor greater than 1, then # $C$ also shares that common factor? Yes, because if $p$ is a factor of both $A$ and $B$, then it is a factor of $A^x + B^y$, and since we know this is equal to $C^z$, then $p$ must also be a factor of $C^z$, and thus a factor of $C$. # # # # # Faster Arithmetic (mod *p*) # # Arithmetic is slow with integers that have thousands of digits. If we want to explore much further, we'll have to make the program more efficient. An obvious improvement would be to do all the arithmetic module some prime number $p$ that fits in one word. Then we know: # # $$\mbox{if} ~~ # A^x + B^y = C^z # ~~ \mbox{then} ~~ # A^x (\mbox{mod} ~ p) + B^y (\mbox{mod} ~ p) = C^z \;(\mbox{mod} ~ p)$$ # # # So we can do efficient tests modulo $p$, and then do the full arithmetic only for combinations that work modulo $p$. Unfortunately there will be collisions (two numbers that are distinct, but are equal mod $p$), so the tables will have to have lists of values. Here is a simple, unoptimized implementation: # In[16]: from math import gcd from itertools import combinations, product from collections import defaultdict def beal_modp(max_A, max_x, p=2**31-1): """See if any A ** x + B ** y equals some C ** z (mod p), with gcd(A, B) == 1. If so, verify that the equation works without the (mod p). Consider any 1 <= A,B <= max_A and x,y <= max_x, with x,y prime or 4.""" assert p >= max_A Apowers = make_Apowers_modp(max_A, max_x, p) Czroots = make_Czroots_modp(Apowers) for (A, B) in combinations(Apowers, 2): if gcd(A, B) == 1: for (Axp, x), (Byp, y) in product(Apowers[A], Apowers[B]): Czp = Axp + Byp if Czp in Czroots: lhs = A ** x + B ** y for (C, z) in Czroots[Czp]: if lhs == C ** z: print('{} ** {} + {} ** {} == {} ** {} == {}' .format(A, x, B, y, C, z, C ** z)) def make_Apowers_modp(max_A, max_x, p): "A dict of {A: [(A**3 (mod p), 3), (A**4 (mod p), 4), ...]}." exponents = exponents_upto(max_x) return {A: [(pow(A, x, p), x) for x in (exponents if (A != 1) else [3])] for A in range(1, max_A+1)} def make_Czroots_modp(Apowers): "A dict of {C**z (mod p): [(C, z),...]}" Czroots = defaultdict(list) for A in Apowers: for (Axp, x) in Apowers[A]: Czroots[Axp].append((A, x)) return Czroots # Here we see that each entry in the `Apowers` table is a list of `(A**x (mod p), x)` pairs. # For example, $6^7 = 279,936$, so in our (mod 1000) table we have the pair `(936, 7)` under `6`. # In[17]: Apowers = make_Apowers_modp(6, 10, 1000) Apowers # And each item in the `Czroots` table is of the form `{C**z (mod p): [(C, z), ...]}`. # For example, `936: [(6, 7)]`. # In[18]: make_Czroots_modp(Apowers) # Let's run the program: # In[19]: get_ipython().run_line_magic('time', 'beal_modp(500, 100)') # This is a bit faster than the previous version, and the idea is that as we start dealing with much larger integers, this version will be even faster, relatively. I could improve this version by caching certain computations, managing the memory layout better, moving some computations out of loops, considering using multiple primes (as in a Bloom filter), finding a way to parallelize the program, and re-coding in a faster compiled language (such as C++ or Go or Julia). Then I could invest thousands (or millions) of CPU hours searching for counterexamples. # # But [Witold Jarnicki](https://plus.sandbox.google.com/+WitoldJarnicki/posts) and [David Konerding](http://www.konerding.com/~dek/) already did that: they wrote a C++ program that built a table of $C^z \;(\mbox{mod} \; p)$ up to $5000^{5000}$, and, in parallel across thousands of machines, searched for $A, B$ up to 200,000 and $x, y$ up to 5,000, but found no counterexamples. On a smaller scale, Edwin P. Berlin searched all $C^z$ up to $10^{17}$ and also found nothing. So I don't think it is worthwhile to continue on that path. # # Conclusion # # This was fun, but I can't recommend anyone spend a serious amount of computer time looking for counterexamples to the Beal Conjecture—the money you invest in computer time would be more than the expected value of your prize winnings. I suggest you work on a proof rather than a counterexample, or work on some other interesting problem instead!