1. Introduction
The nth s-gonal number, with
$s \geq 3$
, which we denote by
$\mathcal {P}_s(n)$
, is given by the formula
$$ \begin{align*} \mathcal{P}_s(n) = \frac{(s-2)n^2-(s-4)n}{2}. \end{align*} $$
Polygonal numbers have been studied since antiquity [Reference Dickson6, pages 1–39] and relations between different polygonal numbers and perfect powers have received much attention (see, for example, [Reference Kim, Park and Pintér7] and the references cited therein). Kim et al. [Reference Kim, Park and Pintér7, Theorem 1.2] found all solutions to the equation
$\mathcal {P}_s(n) = y^m$
when
$m>2$
and
$s \in \{3,5,6,8,20 \}$
for integers n and y. We extend this result (for
$m>1$
) to the case
$s=10$
, that of decagonal numbers.
Theorem 1.1. All solutions to the equation
$$ \begin{align} \mathcal{P}_{10}(n) = y^m,\quad n,y,m \in \mathbb{Z},\quad m> 1 \end{align} $$
satisfy
$n=y=0, n=|y|=1$
or
$n=y=m=3$
.
In particular, the only decagonal number greater than 1 expressible as a perfect mth power with
$m>1$
is
$\mathcal {P}_{10}(3) = 3^3$
.
We will prove Theorem 1.1 by carrying out a descent argument to obtain various ternary Diophantine equations, to which one may associate Frey elliptic curves. The difficulty in solving the equation
$\mathcal {P}_{s}(n) = y^m$
for a fixed value of s is due to the existence of the trivial solution
$n=y=1$
(for any value of m). We note that adapting our method of proof also works for the cases
$s \in \{3,5,6,8,20 \}$
mentioned above, but will not extend to any other values of s (see Remark 3.1).
2. Descent and small values of m
We note that it will be enough to prove Theorem 1.1 in the case
$m=p$
, prime. We write (1.1) as
$$ \begin{align} n(4n-3) = y^p, \quad n,y \in \mathbb{Z},\quad p \text{ prime}\end{align} $$
and suppose that
$n,y \in \mathbb {Z}$
satisfy this equation with
$n \ne 0$
.
Case 1:
$3 \nmid n$
If
$3 \nmid n$
, then n and
$4n-3$
are coprime, so there exist coprime integers a and b such that
$$ \begin{align*} n = a^p \quad \text{ and } \quad 4n-3 = b^p. \end{align*} $$
It follows that
$$ \begin{align} 4a^p-b^p = 3. \end{align} $$
If
$p=2$
, we see that
$(2a-b)(2a+b)=3$
, so that
$a = b = \pm 1$
and so
$n=|y|=1$
. If
$p=3$
or
$p=5$
, then using the Thue equation solver in Magma [Reference Bosma, Cannon and Playoust5], we also find that
$a=b=1$
.
Case 2:
$3 \parallel n$
Suppose that
$3 \parallel n$
(that is,
$\mathrm {ord}_3(n)=1$
). Then, after dividing (2.1) by
$3^{\mathrm {ord}_3(y)p}$
, we see that there exist coprime integers t and u with
$3 \nmid t$
such that
$$ \begin{align*}n = 3t^p \quad \text{ and } \quad 4n-3 = 3^{p-1}u^p. \end{align*} $$
Then
$$ \begin{align} 4t^p-3^{p-2}u^p = 1. \end{align} $$
If
$p=2$
, we have
$(2t-u)(2t+u)=1$
, which has no solutions. If
$p=3$
, then we have
$4t^3-3u^3=1$
and, using the Thue equation solver in Magma [Reference Bosma, Cannon and Playoust5], we verify that
$u=t=1$
is the only solution to this equation. This gives
$n=y=3$
. If
$p=5$
, Magma’s Thue equation solver shows that there are no solutions.
Case 3:
$3^2 \mid n$
If
$3^2 \mid n$
, then
$3 \parallel 4n-3$
and, arguing as in Case 2, there exist coprime integers v and w with
$3 \nmid w$
such that
$$ \begin{align*}n = 3^{p-1}v^p \quad \text{ and } \quad 4n-3 = 3w^p. \end{align*} $$
So,
$$ \begin{align} 4 \cdot 3^{p-2}v^p- w^p = 1. \end{align} $$
If
$p=2$
, then as in Case 2 we obtain no solutions. If
$p=3$
or
$p=5$
, then we use Magma’s Thue equation solver to verify that there are no solutions with
$v \ne 0$
.
3. Frey curves and the modular method
To prove Theorem 1.1, we will associate Frey curves to equations (2.2), (2.3) and (2.4) and apply Ribet’s level-lowering theorem [Reference Ribet8, Theorem 1.1] to obtain a contradiction. We describe this process as level-lowering the Frey curve. We have considered the cases
$p=2,3$
and
$5$
in Section 2 and so we will assume that
$m=p$
is prime with
$p \geq 7$
.
We note that at this point we could directly apply [Reference Bennett and Skinner3, Theorem 1.2] to conclude that the only solutions to (3.1) are
$a=b=1$
, giving
$n=1$
, and apply [Reference Bennett2, Theorem 1.2] to show that (3.2) and (3.3) have no solutions. The computations for (3.1) are not explicitly carried out in [Reference Bennett and Skinner3], so for the convenience of the reader and to highlight why the case
$s=10$
is somewhat special, we provide some details of the arguments.
Case 1:
$3 \nmid n$
We write (2.2) as
$$ \begin{align} -b^p +4a^p = 3 \cdot 1^2, \end{align} $$
which we view as a generalised Fermat equation of signature
$(p,p,2)$
. We note that the three terms are integral and coprime.
We suppose that
$ab \ne \pm 1$
. Following the recipes of [Reference Bennett and Skinner3, pages 26–31], we associate Frey curves to (3.1). We first note that b is odd, since
$b^p = 4n-3$
. If
$a \equiv 1 \pmod {4}$
, we set
$$ \begin{align*}E_1: Y^2 = X^3 -3X^2+3a^pX. \end{align*} $$
If
$a \equiv 3 \pmod {4}$
, we set
$$ \begin{align*}E_2: Y^2 = X^3 + 3X^2+3a^pX. \end{align*} $$
If a is even, we set
$$ \begin{align*}E_3: Y^2+XY = X^3 - X^2 + \frac{3a^p}{16} X. \end{align*} $$
We level-lower each Frey curve and find that for
$i=1,2,3,$
we have
$E_i \sim _p f_i$
for
$f_i$
a newform at level
$N_{p_i}$
, where
$N_{p_1} = 36, N_{p_2} = 72$
and
$N_{p_3} = 18$
. The notation
$E \sim _p f$
means that the mod-p Galois representation of E arises from f. There are no newforms at level
$18$
and so we focus on the curves
$E_1$
and
$E_2$
. There is a unique newform,
$f_1$
, at level
$36$
, and a unique newform,
$f_2$
, at level
$72$
.
The newform
$f_1$
has complex multiplication by the imaginary quadratic field
$\mathbb {Q}(\!\sqrt {-3})$
. This allows us to apply [Reference Bennett and Skinner3, Proposition 4.6]. Since
$2 \nmid ab$
and
$3 \nmid ab$
, we conclude that
$p=7$
or
$13$
and that all elliptic curves of conductor
$2p$
have positive rank over
$\mathbb {Q}(\!\sqrt {-3})$
. However, it is straightforward to check that this is not the case for
$p=7$
and
$13$
. We conclude that
$E_1 \not \sim _p f_1$
.
Let
$F_2$
denote the elliptic curve with Cremona label 72a2 whose isogeny class corresponds to
$f_2$
. This elliptic curve has full two-torsion over the rationals and has j-invariant
$2^{4} \cdot 3^{-2} \cdot 13^{3}$
. We apply [Reference Bennett and Skinner3, Proposition 4.4], which uses an image of inertia argument, to obtain a contradiction in this case too.
Remark 3.1. The trivial solution
$a=b=1$
(or
$n=y=1$
) corresponds to the case
$i=1$
above. The only reason we are able to discard the isomorphism
$E_1 \sim _p f_1$
is because the newform
$f_1$
has complex multiplication. The modular method would fail to eliminate the newform
$f_1$
otherwise. For each value of s, we can associate to (1.1) generalised Fermat equations of signature
$(p,p,2)$
,
$(p,p,3)$
and
$(p,p,p)$
. We found we could only obtain newforms with complex multiplication (when considering the case corresponding to the trivial solution) when
$s = 3, 6, 8, 10$
or
$20$
. A similar strategy of proof also works for
$s=5$
using the work of Bennett [Reference Bennett1, page 3] on equations of the form
$(a+1)x^n-ay^n = 1$
to deal with the trivial solution.
Case 2:
$3 \parallel n$
We rewrite (2.3) as
$$ \begin{align} 4t^p-3^{p-2}u^p = 1 \cdot 1^3, \end{align} $$
which we view as a generalised Fermat equation of signature
$(p,p,3)$
. The three terms are integral and coprime. We suppose that
$tu \ne \pm 1$
. Using the recipes of [Reference Bennett, Vatsal and Yazdani4, pages 1401–1406], we associate to (3.2) the Frey curve
$$ \begin{align*}E_4: Y^2 + 3XY - 3^{p-2}u^p \, Y = X^3. \end{align*} $$
We level-lower
$E_4$
and find that
$E_4 \sim _p f$
, where f is a newform at level
$6$
, an immediate contradiction, as there are no newforms at level
$6$
.
Case 3:
$3^2 \mid n$
We rewrite (2.4) as
$$ \begin{align} -w^p + 4 \cdot 3^{p-2}v^p = 1 \cdot 1^3, \end{align} $$
which we view as a generalised Fermat equation of signature
$(p,p,3)$
. The three terms are integral and coprime. We suppose that
$vw \ne \pm 1$
. The Frey curve we attach to (3.3) is
$$ \begin{align*} E_5: Y^2 + 3XY + 4 \cdot 3^{p-2} v^p \, Y = X^3. \end{align*} $$
We level-lower and find that
$E_5 \sim _p f$
, where f is a newform at level
$6$
, a contradiction as in Case 2.
This completes the proof of Theorem 1.1.
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