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Beal's conjecture is a generalization of Fermat's last theorem. Fermat's last theorem states that there are no solutions to the equation $A^N+B^N=C^N$ where $A,B,C,N\in \Bbb{N}\space |\space N\ge 3$ In Beal's conjecture the exponents are allowed to vary independently, but there is an added restriction that the bases are coprime. So Beal's conjecture states that there are no solutions to the equation $A^X+B^Y=C^Z$ where $A,B,C,X,Y,Z\in\Bbb{N}\space |\space X,Y,Z \ge 3$ and $gcd(A,B)=gcd(B,C)=gcd(A,C)=1$

(As far as I'm aware this has not been proven or disproven yet.)

At one point Ramanujan looked into Fermat's last theorem he created generating functions which gave an infinite number of near miss solutions to Fermat's last theorem they are of the form $A^3+B^3=C^3\pm 1$ here are the generating functions and a few examples

My question: is there at least one near miss solution to Beal's conjecture? $$A^X+B^Y=C^Z\pm 1 \quad \text{{1}}$$ where $A,B,C,X,Y,Z\in\Bbb{N}\space |\space X,Y,Z \ge 3$ and $gcd(A,B)=gcd(B,C)=gcd(A,C)=1$

To be clear one near miss solution is sufficient. A generating function isn't necessary. (Although it would be extremely cool if someone finds one.)

Every equation which consists only of integer terms has an even number of odd terms. There can't be zero odd terms in {1} because $\pm 1$ is odd. There can't be two odd terms in {1} because that implies that there are two even terms and therefore one of the three gcd pairs isn't one. This leaves the only possibility that all terms in {1} are odd. Only odd bases can generate odd power numbers so $A$,$B$, and $C$ are all odd.

I have found two cases where all conditions are met except for coprime condition $3^3+15^4=37^3-1$ and $7^4+19^3=21^3-1$

I also found another case where the power numbers meet all conditions except for the $\pm 1$ and was only two off $31^3+63^3=23^4-3$

If there is a counter example one of the terms in {1} must be above $10^6$

Edit: I Realize that I should have stated this from the beginning. I want to avoid really trival counter examples so $A\neq 0, B\neq 0, C\neq 0, C\neq 1$. One of $A$ or $B$ can equal one but if this is the case the equation of the form $A^X+B^Y=C^Z+1$ cannot be used. If one of $A$ or $B$ is one then only the form $A^X+B^Y=C^Z-1$ is allowed. (If one of $A$ or $B$ is one then this would be equivalent to finding a pair of odd power numbers where both of the exponents are greater than two and the difference between the numbers in the pair is two. I don't think such a pair exists.)

Cotton Headed Ninnymuggins
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quantus14
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    Rational points near curves and small nonzero |x^3-y^2| via lattice reduction by Noam D. Elkies, https://arxiv.org/abs/math/0005139 might be of interest. – Gerry Myerson Jun 10 '19 at 00:50
  • $\gcd(x,y,z)\leq 2$ otherwise we run into a subset of Fermat's last theorem. –  Jun 10 '19 at 11:31
  • @RoddyMacPhee for Beal's conjecture this is true. For my near miss problem not necessarily. – quantus14 Jun 10 '19 at 11:42
  • any near miss with C odd, has A and B same parity, coprimality ensures they are both odd. –  Jun 10 '19 at 11:51
  • @RoddyMacPhee I said this already "This leaves the only possibility that all terms in {1} are odd." $A$,$B$, and $C$ all must be odd to produce odd power terms. – quantus14 Jun 10 '19 at 12:02
  • Another near miss is at $3^7+2^3=13^3-2$. – Servaes Jun 11 '19 at 15:17
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    @Servaes ($3^7 + 2^3 = 13^3 - 2$): Near miss of a near miss of Beal's conjecture. ($31^3+63^3= 23^3 -3$): Near missception of Beal's conjecture. – quantus14 Jun 11 '19 at 19:21
  • A near miss without coprime values is given by $6^3+8^3=3^6-1$, or by the famous $10^3+9^3=12^3+1$. – Servaes Jun 12 '19 at 19:30
  • Also of interest is an amazing identity by Ramanujan that produces infinitely many triples of integers $(a,b,c)$ such that $$a^3+b^3=c^3\pm1,$$ which are near misses to Beal's conjecture, except for the fact that $b$ and $c$ will always both be even. – Servaes Jun 12 '19 at 19:38
  • @Servaes yes the famous taxi cab numbers. Both of these examples are from the generating functions that Ramannujan made which is in the link I have provided. – quantus14 Jun 12 '19 at 19:40
  • @Servaes Interesting link I haven't seen that derivation before for his generating functions. Definitely something worth investigating further. – quantus14 Jun 12 '19 at 19:48
  • There is a near miss in the sense $31 = 4 + 27 = 2^{\color{red}2} + 3^3 = 2^5 - 1$. Here $X < 3$ and does not meet the listed criteria. Perhaps we should check other Mersenne primes... – vvg Oct 29 '20 at 09:28
  • If you allow $Z = 2$, there are a family of solutions here at this link for $$x^4 + y^4 \pm 1 = z^2$$

    See: http://members.bex.net/jtcullen515/math10.htm

     – vvg Oct 29 '20 at 09:44
  • $6^3 + 8^3 = 6^3 + 2^9 = 3^6 - 1$. But $A,B$ are not co-prime. – vvg Oct 29 '20 at 10:15

1 Answers1

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Among the sixty solutions of $A^3 + B^3 = C^3 - 1$ tabulated in OEIS sequences A050787, A050788 and A050789, there are three that also meet the coprimality condition: $$ 25765^3 + 33857^3 = 38239^3 - 1, $$ $$ 54101^3 + 56503^3 = 69709^3 - 1, $$ and $$ 51293^3 + 64165^3 = 73627^3 - 1. $$ In the second example, each of $A,B,C$ is actually a prime.

Similarly, $368$ solutions of $A^3 + B^3 = C^3 + 1$ are tabulated in OEIS sequences A050791, A050792 and A050793. In this much longer list, $18$ triples are pairwise coprime; here are the first three:

$$ 35131^3 + 76903^3 = 79273^3 + 1, $$ $$ 190243^3 + 219589^3 = 259495^3 + 1, $$ and $$ 466727^3 + 867679^3 = 910541^3 + 1. $$ Only one of the $18$ solutions in this set involves three primes: $$ 457047089^3 + 552546031^3 = 641644319^3 + 1. $$

FredH
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  • (As you may have already seen) I have a link that shows a generating function that Ramanujan created that gives infinitely many near miss solutions to Fermat's last theorem. These solutions are a small subset of the solutions you mentioned in the link to OEIS. I assumed that none of the solutions that come from the generating function were coprime because the specific examples shown were not coprime and were produced in a particular pattern. What you have shown me is that this may not be the case. Assuming that near miss solutions are random and knowing that the probability of any ... – quantus14 Nov 02 '20 at 02:11
  • two random coprime natural numbers are $\frac{6}{\pi^2}$ then the probability that a triplet solution is coprime is $\left(\frac{6}{\pi^2}\right)^3\approx 0.225$. So I not only believe that there are infinitely many solutions to the problem I posed, but also that these solutions are relatively common. (+1) – quantus14 Nov 02 '20 at 02:12