TITLE: Fermat-like equation $c^n=a^{2n}+a^n b^n + b^{2n}$
QUESTION [13 upvotes]: Hello,
Is there a solution to the following equation:
$c^n=a^{2n}+a^n b^n + b^{2n}$
where $a,b,c \in \mathbb{N}^*$, and $n$ is an integer $\geq 2$.
The problem is due to Antoine Balan.
Thanks in advance.

REPLY [12 votes]: Yes, it's a nice question.  It does seem that there are no solutions in
positive integers to A. Balan's equation, and indeed no integer
solutions at all other than those with $a=0$, $b=0$, or (when $n$ is
odd) $a+b=0$.  As usual the problem is equivalent to finding all
rational points on an algebraic curve $B_n$; here $B_n$ is the
"superelliptic" curve with affine equation $y^n = x^{2n} + x^n + 1$.
We want to show that the only rational points are those with
$x=\infty$, $x=0$, and (when $n$ is odd) $x=-1$.  Herewith:
1) Verification that $ab$ and $c$ may be assumed coprime; by
M.Bennett's clever rewriting and the results he quotes from DMP =
Darmon, Merel, and Poonen, this reduces the problem to $n=2$ and $n=3$.
2) $n=2$, worked out in more detail than in my comment to the original
question.
3) $n=3$; fortunately a genus-2 quotient $C_3$ of $B_3$ has torsion Jacobian.
4) A bit more on the analogy JSE suggested:
Balan is to $\lbrace0,\infty,\rho,\rho^2 \rbrace$
as Fermat is to $\lbrace 0, 1, \infty \rbrace$
(where $\rho = e^{2\pi i/3}$), and why the proof suggested
by this analogy seems to barely fail.
5) Where Bennett's approach fits in this picture.
(1) Mike finished his answer with the sentence "The condition on
coprimality might be a problem, but it's too early in the morning for
me to be sure!"; by now he's probably had the chance to check this, but
I don't see it yet as an edit or a comment, so here goes.  To apply DMP
to $c^n + (ab)^n = (a^n+b^n)^2$ we need $c$ coprime to $ab$.  If $p$
divides $c$ and (say) $a$, then taking the equation mod $p$ yields
$p|b^n$, and then $p|b$, so $p^{2n} | a^{2n} + a^n b^n + b^{2n} = c^n$,
whence $p^2 | c$.  At this point we may replace $(a,b,c)$ by the equivalent
solution $(a/p, b/p, c/p^2)$.  Unless $(a,b,c)=(0,0,0)$, in finitely
many steps we reach an equivalent solution with $\gcd(ab,c)=1$ as
desired.  If $n\geq 4$, it follows from DMP that at least one of $ab$,
$a^n+b^n$, and $c$ vanishes, and we're done.
It remains to deal with $n < 4$.
(2)  $B_2: y^2 = x^4 + x^2 + 1$ is an elliptic curve.  It has
rational points at infinity, so we put it in extended Weierstrass form
by the usual centuries-old technique.  Choose a point $P$ at infinity,
and expand $y$ about $P$ in a Laurent series, truncating before the
constant term to get a quadratic polynomial; here it's just $x^2$.
Hence $t := y+x^2$ has a double pole at $P$ and nowhere else on the curve,
so it can be used as the abscissa.  Writing $y=t-x^2$ in $y^2 = x^4 + x^2 + 1$
gives a quadratic in $x$, namely $(2t+1)x^2 = t^2-1$, which has solutions
iff $4(2t+1)(t^2-1)=u^2$ for some $u$;
dividing $t$ by $2$ yields $(t+1)(t-2)(t+2) = u^2$, which is
the curve of conductor $48$ with coefficient vector $[0,1,0,-4,-4]$.
Thanks to rational $2$-torsion (and trivial Sha[2] obstruction),
Fermat's descent technique, now automated in Cremona's mwrank,
suffices to prove the curve has rank zero and no rational points
other than the four trivial ones we already knew of.
(3)  $B_3: y^3 = x^6 + x^3 + 1$ has genus $4$.  There's an obvious
quotient curve of genus $1$, namely
$E: Y^{\phantom.3} = X^{\phantom.2} + X + 1$,
which is already in the usual extended Weierstrass form
(with the roles of $X$ and $Y$ reversed).  Unfortunately the
obvious points with $Y=1$ are of infinite order; indeed plugging the
coefficient vector $[0,0,1,0,-1]$ into mwrank we find that
$(X,Y\phantom.) = (0,1)$ (or equivalently $(-1,1)$)
generates the group of rational points.  (We could have known in advance
that the rank cannot exceed $1$ because its conductor is $3^5 = 243 < 389$...)
So we cannot use this quotient to prove that $B_3$ has no more rational points.
Fortunately there is a genus-2 quotient $C_3: Y^{\phantom.3} = X^4 + X^3 + X^2$
that does work.  [To get this quotient, first write $B_3$ as
$(x^2 y)^3 = x^{12} + x^9 + x^6$, then take $(X,Y\phantom.) = (x^2 y, x^3)$.]
To put this curve in hyperelliptic form, divide through by $X^2$
to get $z^3 X = X^2 + X + 1$ where $z = Y/X$, then find the discriminant
of this quadratic in $X$ to get $z^6 - 2z^3 - 3 = w^2$.  The three
known points correspond to the Weierstrass point $(z,w) = (-1,0)$
and the two points at infinity, call them $P_\pm$.
The subgroup of the Jacobian $J(C_3)$ represented by degree-zero
divisors supported on these three points is isomorphic with
${\bf Z} / 6 {\bf Z}$: twice the Weierstrass point is equivalent with
$P_+ + P_-$, and $3P_+ \sim 3 P_-$ because the difference is the
divisor of $z^3 - 1 - w$.  [An analogous result holds for all odd $n$,
but the next step might not.]  It turns out that this torsion group
is all of $(J(B_1))({\bf Q})$: I asked magma
_<z> := PolynomialRing(Rationals());
C1 := HyperellipticCurve(z^6 - 2*z^3 - 3);
J := Jacobian(C1);
TorsionSubgroup(J);
RankBound(J);

(adapting one of the examples in
http://magma.maths.usyd.edu.au/magma/handbook/text/1375),
and found that the Jacobian $J(C_3)$ has $6$ torsion points and rank zero.
It follows that our list of rational points on $C_3$, and thus
also on $B_3$, is complete.
[The Jacobian of the genus-$4$ curve $B_3$ is isogenous with
$E \times E \times J(C_3)$: in general the Jacobian of a superelliptic
curve of the form $y^m = P(x^m)$ is isogenous with the product of the
Jacobians of the $n$ curves $Y^{\phantom.m} = X^k P(X)$ with $k=0,1,2,\ldots,m$,
and here $k=0$ and $k=1$ both yield $E$, while $k=2$ yields $C_3$.]
(4)  Recall the following geometrical description of the Frey curves
associated with a putative point on the Fermat curve $F_p: x^p + y^p = z^p$,
with $p>3$ prime.
The rational function $(x/z)^p$ on $F_p$ realizes $F_p$ as a $\mu_p^2$
cover of ${\bf P}^1$, branched only above $\lbrace 0, 1, \infty \rbrace$.
Use the modular function $\lambda$ to identify ${\bf P}^1$ with
the modular curve ${\rm X}(2)$, whose three cusps are at $\lambda=0,1,\infty$.
The image of a putative Fermat counterexample yields a Frey curve
$E: Y^{\phantom.^2} = X (X \pm x^p) (X \mp y^p)$,
with the sign chosen to make the curve semistable even at $2$
(this is where $p>3$ is needed), and thus modular.
But the cover ${\rm X}(2p) \rightarrow {\rm X}(2)$ of modular curves
has the same ramification points and orders as our map from $F_p$.
Once $E$, and thus the Galois representation on $E[p]$, has been proved modular,
it then follows that this mod-$p$ representation comes from a cuspform
of level $2$, which is impossible because fortunately ${\rm X}_0(2)$
has genus zero.
Now the Balan curve $B_p$ is likewise a $\mu_p^2$ cover of ${\bf P}^1$,
but with four branch points, at $(a/b)^p = 0$, $\infty$, $\rho$, and $\rho^2$.
These don't seem to be the cusps of any modular curve.  But for our purpose
it's enough for some subset to be the cusps, and JSE suggested the only
possibility: twist $\Gamma(2)$ by ${\bf Q}(\sqrt{-3})$ to
put two of its cusps at $\rho$ and $\rho^2$ and the third at $\infty$
(or equivalently at $0$).  Then the Frey curve becomes
$$
E': Y^{\phantom.2} = X \bigl(X^2 + (a^p+2b^p) X + (a^{2p}+a^pb^p+b^{2p}) \bigr),
$$
with discriminant $-48 a^{2p} (a^{2p}+a^pb^p+b^{2p})^p = -48 a^{2p} c^p$.
So we must deal with reduction at $3$ as well as at $2$.  Now the
good news is that $E'$ is semistable at $3$: while there are coprime
integers $A,B$ such that
$Y^{\phantom.2} = X \bigl(X^2 + (A+2B) X + (A^2+AB+B^2)\bigr)$
has additive reduction at $3$, this can only happen if $A\equiv B \mod 3$,
but then $A^2 + AB + B^2 \equiv 3 \bmod 9$ so it cannot be a $p$-th power.
The bad news is that $E'$ is not in general semistable at $2$,
and even after applying the available symmetries $(a,b) \leftrightarrow (b,a)$
and $(a,b) \leftrightarrow (-a,-b)$ the power of $2$ in the conductor
might be as large $2^3$ $-$ this happens exactly when $a \equiv b \bmod 4$.
In this case, all that we can say is that the Galois representation
on $E'[p]$ is isomorphic to the the $p$-torsion of an elliptic curve
of conductor 24, such as $y^2 = x^3 - x^2 + x$ (it is not entirely
coincidental that this curve is a quadratic twist of the one we
encountered above for exponent $2$).  At this point I'm stuck:
there might be a way to push this argument further or modify it
to finish off the proof, but all I can obtain this way is the
unsatisfying conclusion that in any primitive solution $a \equiv b \bmod 4$
and $E'[p]$ has the same Galois structure as the $p$-torsion of
$y^2 = x^3 - x^2 + x$.  The first condition follows from the computation
that in all other cases $E'[p]$ has conductor dividing $12$ for at least one
of the four equivalent choices of $(a,b)$, and then there's no possible
cuspform because ${\rm X}_0(12)$ is rational.  This condition is peculiar
because the curve $B_n$ has good reduction at 2 (it's the construction of
the auxiliary curve $E'$ introduces problems at this prime).  The
condition on $E'[p]$ seems rather hard to exploit...
(5) Fortunately there's another way, which comes down to what
Bennett found: compose the map, call it
$t = (a/b)^p: B_p \rightarrow {\bf P}^1$,
with the map $s: t \mapsto t + 1/t$, giving the quotient map
${\bf P}^1 \rightarrow {\bf P}^1$ under the involution
$t \leftrightarrow 1/t$ that switches $0$ with $\infty$ and
$\rho$ with $\rho^2$.  This map has double points at $s = \pm 2$
(images of the fixed points $t = \pm 1$ of the involution),
and takes $t=0,\infty$ to $s=\infty$ and $t = \rho,\rho^2$
to $s = -1$.  So now, in place of ${\rm X}(2)$ we need a
modular curve with cusps at $s=-1$ and $s=\infty$
and an elliptic point of order $2$ at $s=2$ or $s=-2$
(or possibly both).  The modular curve ${\rm X}_0(2)$,
with two cusps and one elliptic point of order $2$,
does the trick, and if we choose to put the elliptic point
at $s=-2$ then there's no problem with reduction at $3$
(putting it at $s=+2$ would cause such a problem because
$2 \equiv -1 \bmod 3$).  This yields the elliptic curve
$E'': Y^2 = X \bigl(X^2 \pm 2(a^p+b^p) X + c^p \bigr)$.
That's still not the end of the game: the smallest
valuation at $2$ of any quadratic twist of $E''$ is $5$,
and the modular curve ${\rm X}_0(2^5)$ has genus $1$,
so there is a modular form of level $2^5$ that must be
dealt with.  Fortunately this form, unlike the one of level $24$,
is CM.  Thus the corresponding Galois representation mod $p$
is well understood, and with some further cleverness and much effort
Darmon and Merel were able to dispose of this final case.