gpt 6 astra ultra attempt on Math-319

Fermat-Catalan Conjecture

Definitions and mathematical statement

For positive integers $x,y,z$ and integers $p,q,r>1$, consider \[ x^p+y^q=z^r,\qquad \gcd(x,y)=\gcd(y,z)=\gcd(z,x)=1, \qquad\frac1p+\frac1q+\frac1r<1. \] The recorded finiteness assertion concerns the set of base triples $(x,y,z)$ that admit at least one such exponent triple: this set is finite. The strict reciprocal-sum condition excludes the spherical and Euclidean exponent ranges. Ordering of $x,y$ is harmless for finiteness, but positivity and the coprimality restriction are essential. The exact source wording counts base triples; this must be distinguished from versions of Fermat--Catalan that count triples of powers $(x^p,y^q,z^r)$ or full sextuples. This entry does not silently identify those counting conventions.

Formulation and concept references: [S1].

Short English statement

Are there only finitely many coprime positive base triples whose powers satisfy an additive equation with reciprocal exponents summing to less than one?

Research attempt

Status: UNRESOLVED. The distinction between base triples, power triples and full exponent data is made explicit. A uniform reciprocal-exponent gap gives a complete conditional deduction from the usual $abc$ conjecture. Unit-base solutions and fixed signatures are separated from the unrestricted union of signatures. No unconditional bound for that union is obtained.

Source audit and the object being counted.

The lecture notes [S1], Conjecture 11.24 on printed page 151, were inspected on 27 September 2026. They state the coprime hyperbolic power equation and give examples including $1+2^3=3^2$, $2^5+7^2=3^4$ and $7^3+13^2=2^9$. The catalogue explicitly selects finiteness of base triples, and this attempt retains that convention. The classical paper of Darmon and Granville [E1] supplies an external finiteness theorem for each fixed hyperbolic signature; its conclusion is not uniform in the signature.

Write $A=x^p$, $B=y^q$, $C=z^r=A+B$. Positivity forces $z\ge2$. If a prime divides two of $x,y,z$, the equation forces it to divide the third. Consequently, for solutions of this equation, the pairwise coprimality condition is equivalent to $\gcd(x,y,z)=1$. It also implies pairwise coprimality of $A,B,C$.

1. Why full sextuples cannot be the selected finite set.

For every integer $p\ge7$, \[ 1^p+2^3=3^2,\qquad \frac1p+\frac13+\frac12<1. \] This produces infinitely many sextuples $(1,2,3,p,3,2)$, but only one base triple and one power triple. Thus an unqualified finiteness claim about all sextuples allowing base one would already be false. The catalogue avoids that problem by counting bases.

If all bases are at least two and a power bound $C\le H$ is available, then each base is at most $\sqrt H$, and each exponent is at most $\log_2 H$. Therefore a bound for $C$ yields finiteness of base triples and also of the full nonunit sextuples. With a unit base, its exponent can still vary without changing the power. These are elementary one-way implications; no converse is presumed merely from the terminology of the conjecture.

2. The complete unit-base subcase using an external theorem.

If $x=1$, then $z^r-y^q=1$. If $y=1$ as well, this would give $z^r=2$, impossible for $r>1$. Otherwise the theorem on consecutive perfect powers proved by Mihăilescu, stated in [S1], gives $y^q=8$ and $z^r=9$. The only allowed representations of these powers are $y=2,q=3$ and $z=3,r=2$. The hyperbolic inequality then requires $p\ge7$.

The case $y=1$ is symmetric, and $z=1$ is impossible by positivity. Hence the only base triples with a unit entry are $(1,2,3)$ and $(2,1,3)$. This conclusion uses the full consecutive-perfect-powers theorem as an imported input; it is not a new elementary proof of that theorem. From this point one can restrict attention to $x,y,z\ge2$.

3. A uniform gap below one for reciprocal sums.

Let $a\le b\le c$ be a reordering of $p,q,r$, solely for estimating their reciprocal sum. If that sum is less than one, then \[ \frac1p+\frac1q+\frac1r\le\frac{41}{42}. \] Here is an exhaustive proof. If $a\ge4$, the sum is at most $3/4$. If $a=3$ and $b\ge4$, it is at most $1/3+1/4+1/4=5/6$. If $a=b=3$, the strict inequality forces $c\ge4$, giving at most $11/12$.

Now let $a=2$. The choice $b=2$ is impossible. If $b=3$, the strict inequality forces $c\ge7$, giving the maximum $1/2+1/3+1/7=41/42$. If $b=4$, it forces $c\ge5$, giving at most $19/20$. If $b\ge5$, the sum is at most $1/2+2/5=9/10$. These cases exhaust all triples. Equality in the largest bound occurs exactly for permutations of $(2,3,7)$.

This discreteness is important. Merely saying the reciprocal sum is less than one would leave an exponent-dependent gap in an $abc$ argument. The uniform gap $1/42$ permits a single choice of $abc$ parameter for every signature.

4. A complete conditional consequence of $abc$.

Assume the standard positive-integer $abc$ statement: for each $\varepsilon>0$ there is a constant $K_\varepsilon$ such that every pairwise coprime $A+B=C$ satisfies \[ C\le K_\varepsilon\operatorname{rad}(ABC)^{1+\varepsilon}, \] where the radical is the product of the distinct prime divisors. Since $A,B<C$ and $C=z^r$, \[ x\le C^{1/p},\qquad y\le C^{1/q},\qquad z=C^{1/r}, \qquad \operatorname{rad}(ABC)=\operatorname{rad}(xyz)\le xyz \le C^{41/42}. \] Choose the fixed value $\varepsilon=1/83$. Then \[ \frac{41}{42}\left(1+\frac1{83}\right)=\frac{82}{83}, \qquad C\le K_{1/83}C^{82/83}. \] Thus $C\le\max(1,K_{1/83})^{83}$. This is a bound independent of $p,q,r$. There are only finitely many positive integer power triples with $A+B=C$ below it, and therefore only finitely many base triples. This proves the selected conjecture conditionally on the stated $abc$ inequality.

The deduction does not prove $abc$, does not make its constant effective, and does not prove that a proposed finite list is complete. Even an ineffective finite bound gives abstract finiteness, but a claimed exhaustive computation would additionally require an explicit justified bound.

5. Fixed signatures versus their infinite union.

The external Darmon--Granville theorem [E1], using Faltings' theorem, gives finitely many primitive integer solutions for fixed exponents satisfying $1/p+1/q+1/r<1$. Consequently, if the exponents are all bounded by a fixed integer $B$, the selected set of base triples is finite: there are only finitely many signatures to consider.

This does not settle the problem with unbounded exponents. A countable union of finite sets can be infinite, and neither the sizes nor the heights supplied by the fixed-signature theorem have been bounded uniformly here. A useful way to state what a counterexample to finiteness would entail is this: an infinite sequence of distinct base triples must escape every bounded set of signatures, so along a subsequence at least one exponent tends to infinity. This reformulation identifies a uniformity problem; it does not exclude the escaping sequence.

Likewise, fixing the bases and varying exponents is a different family of exponential Diophantine equations. No theorem about that family is assumed to turn finite base counting into finite full data. The explicit unit-base example already prevents an unrestricted identification of the two conventions.

6. Exponent reduction can leave the hyperbolic range.

If a common integer $d\ge3$ divides $p,q,r$, then \[ (x^{p/d})^d+(y^{q/d})^d=(z^{r/d})^d, \] contradicting Fermat's Last Theorem for positive bases. This is another useful imported exclusion. If the common divisor is only two, the reduced equation is Pythagorean and that theorem does not apply.

It is also invalid to replace each exponent by an arbitrary prime divisor and assume the reciprocal inequality survives. For example the signature $(4,4,3)$ has reciprocal sum $5/6<1$, while its componentwise reduction $(2,2,3)$ has sum $4/3>1$. The new bases absorb the discarded exponent factors, so an algebraic equation is retained, but the hypothesis of the fixed-hyperbolic-signature theorem is lost. Any proposed reduction to finitely many signatures must verify both its divisibility bookkeeping and preservation of the needed inequality.

The strict boundary is essential. Primitive Pythagorean triples give infinitely many solutions with signature $(2,2,2)$, outside the target. Other signatures with reciprocal sum at least one have their own arithmetic behavior. An argument that never uses the strict reciprocal condition would need to explain why it does not incorrectly rule out those infinite families.

7. Exact examples, a finite search and the unresolved step.

The displayed nonunit examples can be checked directly: \[ 2^5+7^2=32+49=81=3^4, \qquad 7^3+13^2=343+169=512=2^9. \] The bases are pairwise coprime and the reciprocal sums are strictly below one. Thus the target is finiteness, not absence of solutions. The conditional argument above also allows such examples.

The offline script enumerates powers with bases from 2 through 100 and exponents from 2 through 10, matches sums exactly against the same power dictionary, and retains only pairwise coprime hyperbolic solutions. It records all matches with their exponents rather than inferring an exponent from a floating logarithm. Unit-base identities are checked separately. Rational arithmetic verifies the reciprocal-gap case bounds and exhaustively checks sorted signatures through exponent 80 as a diagnostic of the case analysis.

The search has explicit finite base and exponent cutoffs and does not claim completeness beyond them. The unconditional obstacle is still to control solutions uniformly as signatures escape every finite set. The $abc$ calculation shows exactly one sufficient global inequality; the fixed-signature theorem and the finite search do not supply that inequality. The general finiteness assertion remains unresolved here.

Sources

• S1 Violetta Weger, Elementary Number Theory, Conjecture 11.24 $2024/25$. https://home.cit.tum.de/~wvi/NT.pdf. Formulation source.
• E1 Henri Darmon and Andrew Granville, On the equations $z^m=F(x,y)$ and $Ax^p+By^q=Cz^r$, Bulletin of the London Mathematical Society 27 $1995$, 513--543. https://www.math.mcgill.ca/darmon/pub/Articles/Research/12.Granville/pub12.pdf. External finiteness theorem for a fixed hyperbolic signature, not a uniform theorem over all signatures.