On the Diophantine Equation $p^x+ (2p+1)^y =z^2$ with Consecutive Exponents

Abstract: We study the Diophantine equation $p^x+ (2p+1)^y =z<sup>2$ over positive integers $x$, $y$ and $z$ for every odd prime $p$. We prove that $(x,y,z)=(2,1,p+1)$ is the unique solution except possibly when $2p+1$ is composite. In that case, it reduces to a family depending on one parameter, and we show that the parameter must be odd and satisfies an explicit upper bound, reducing the problem to finitely many cases.

• The paper identifies conditions under which solutions to the Diophantine equation with consecutive exponents are either a unique solution or constrained to specific values.
• One case the paper examines involves factorization of $2p+1$ into coprime, consecutive odd primes; methods used include modular reduction and valuations via the Lifting-the-Exponent Lemma.
• The above factorization is iteratively scaled to fit through logarithmic bound constraints and the equation $ p^x+(2p+1)^y = z^2 $ (scaling factor $(2k^2-1)^{b+1})$.

The paper studies the exponential Diophantine equation

$p^x + (2p+1)^y = z^2$

over positive integers $x$, $y$, $z$, where $p$ is an odd prime and the exponents are required to be consecutive, i.e., $|x-y|=1$ (2608.18608). The novelty relative to earlier work on equations of the form $a^x + b^y = z^2$ is twofold: the pair $(p,\, 2p+1)$ is treated uniformly without assuming that $2p+1$ is prime, and the consecutive-exponent condition—previously examined only for specific pairs such as $(43, 87)$ by Gaha and Mezroui—is handled for the entire one-parameter family.

Main result

The central theorem establishes a dichotomy. For every odd prime $x$0, exactly one of the following holds:

• Case (i): the unique solution in positive integers is

$x$1

which follows from the identity $x$2.

• Case (ii): any exceptional solution can occur only when $x$3 is composite and factors into coprime consecutive odd integers,

$x$4

with the exponents tied to a parameter satisfying

$x$5

The logarithmic bound reduces the problem to finitely many candidate exponents for each fixed prime, converting an infinite family into a decidable finite check per value of $x$6. A concrete exceptional solution exists at $x$7: for $x$8 one has $x$9, i.e., $y$0.

Method of proof

The argument is elementary throughout, relying on coprimality factorizations, modular order arguments, and the Lifting-the-Exponent (LTE) lemma (2608.18608). Since $y$1 and $y$2 are consecutive, they have opposite parity, which splits the problem cleanly.

Case A ($y$3 odd, $y$4 even). Writing $y$5, $y$6 yields $y$7 with $y$8. The two factors are shown to be coprime, so their product being a prime power forces $y$9, giving $z$0. Both parity-consistent sub-cases ($z$1 and $z$2) are then eliminated by reduction modulo $z$3: the congruence forces $z$4, hence $z$5; reduction modulo $z$6 then requires $z$7 via $z$8, contradicting oddness of $z$9. This case therefore admits no solutions with $p$0. The boundary case $p$1 produces only $p$2, which is outside the stated domain since $p$3 must be positive—a small assumption worth noting when comparing with literature that permits $p$4.

Case B ($p$5 even, $p$6 odd). Writing $p$7, $p$8 gives $p$9.

• Trivial split ($|x-y|=1$0): this yields $|x-y|=1$1. For the sub-case $|x-y|=1$2, LTE gives $|x-y|=1$3, so any solution requires $|x-y|=1$4, hence $|x-y|=1$5; an induction shows $|x-y|=1$6 for all $|x-y|=1$7, while $|x-y|=1$8 fails by direct computation. This isolates the unique solution $|x-y|=1$9, namely $a^x + b^y = z^2$0. The sub-case $a^x + b^y = z^2$1 fails by a simple size comparison, $a^x + b^y = z^2$2.
• Non-trivial split: both factors exceed $a^x + b^y = z^2$3, so coprimality forces $a^x + b^y = z^2$4 with $a^x + b^y = z^2$5, $a^x + b^y = z^2$6, and $a^x + b^y = z^2$7. A gcd argument shows $a^x + b^y = z^2$8 where $a^x + b^y = z^2$9, so $(p,\, 2p+1)$0; $(p,\, 2p+1)$1 contradicts integrality of $(p,\, 2p+1)$2, leaving $(p,\, 2p+1)$3, i.e., $(p,\, 2p+1)$4, $(p,\, 2p+1)$5, and $(p,\, 2p+1)$6 prime. Within this structure, the sub-case $(p,\, 2p+1)$7 dies because $(p,\, 2p+1)$8, so only $(p,\, 2p+1)$9 survives, yielding the auxiliary equation above.

The logarithmic bound

The finiteness statement follows from pairing terms in the sum and applying AM–GM:

$2p+1$0

while equation (the auxiliary relation rewritten) requires the same sum to equal $2p+1$1. Dividing through gives

$2p+1$2

since $2p+1$3. Taking logarithms yields $2p+1$4, which is the advertised finite range. Note the bound depends on the primality requirement $2p+1$5 being prime; for composite values of $2p+1$6 the factorization argument does not directly apply.

Limitations and open questions

The paper is explicit about what remains unresolved. Computation for $2p+1$7 shows that no further solutions exist among odd $2p+1$8, suggesting $2p+1$9 may be the sole exception to case (i); however, no proof covering all such cases is given, and the author states plainly that determining whether the auxiliary equation admits solutions beyond $(43, 87)$0 is open. Additionally, the dichotomy is conditional on the coprime-factorization hypothesis in case (ii): if $(43, 87)$1 has a square factor or more than two prime factors arranged differently, the reduction to $(43, 87)$2 would require separate treatment, though the gcd analysis of $(43, 87)$3 and $(43, 87)$4 constrains the possibilities considerably. The exclusion of $(43, 87)$5 also leaves the boundary solution $(43, 87)$6 outside the theorem's scope.

Conclusion

This paper resolves the consecutive-exponent Diophantine equation $(43, 87)$7 up to a single finite auxiliary problem: either the unique explicit solution $(43, 87)$8 holds, or $(43, 87)$9 must be a product of twin-odd factors around $x$00, with exponents confined to $x$01. The techniques are elementary but carefully deployed—LTE valuations, modular order obstructions, and AM–GM size bounds—and the residual question of uniqueness beyond the example $x$02 provides a precise target for subsequent work.