Fermat-like equation $c^n=a^{2n}+a^n b^n + b^{2n}$

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.

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Great question. It is to the set 0,infinity,exp(2pi/3),exp(4pii/3) what Fermat is to (0,infinity,1). I suppose if I wanted to cook up a natural "Frey curve" in this situation I would try to find a way to express that 4-punctured sphere as some friendly (possibly twisted) moduli space of elliptic curves with some kind of level structure. –  JSE Sep 18 '11 at 15:13
I think it is impossible for n=2, since we know that Pythagorean triples are of the form $m^2 - n^2, 2mn, m^2 + n^2$ –  Barinder Banwait Sep 18 '11 at 15:15
Not sure how the impossibility for $n=2$ is supposed to follow from the parametrization of Pythagorean triples as B.Banwait claims. Still it's true that there are no solutions in positive integers with $n=2$. The equation is an elliptic curve isomorphic with $y^2 = (x+1)(x^2-4)$. It has conductor $48$, so we can look it up in tables (or run Cremona's mwrank) to find that there are only the four rational points corresponding to trivial solutions with $ab=0$. –  Noam D. Elkies Sep 18 '11 at 17:05
@JSE: Could you make that analogy more precise? –  Moosbrugger Sep 19 '11 at 19:30

Writing the equation as $$c^n+(ab)^n = (a^n+b^n)^2,$$ one has, via results of Darmon and Merel, and Poonen, that there are no solutions for $n \geq 4$, provided the three terms are coprime. If $n=2$, I think Pythagorean triply stuff shown that there are again no solutions. In the case $n=3$, I guess one could again look at the parametrizations for $X^3+Y^3=Z^2$, say in Cohen's GTM. I'm not sure how hard this case is....

The condition on coprimality might be a problem, but it's too early in the morning for me to be sure!

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SLICK! Amusingly enough, the reason I returned to this question was to add a comment saying "You should really ask Mike Bennett." –  JSE Sep 18 '11 at 17:42
Did not Fermat himself effectively take care of the case n=4k? Gerhard "Ask Me About System Design" Paseman, 2011.09.18 –  Gerhard Paseman Sep 18 '11 at 23:44
Hi Jordan! Gerhard is certainly correct about the cases where $4 \mid n$. The $n=3$ case can likely be attacked using Chabauty, but I don't see it immediately -- unless I'm missing something, the obvious genus one covers have positive rank. –  Mike Bennett Sep 19 '11 at 2:11
Thanks for your answer, Mike. –  user12806 Sep 19 '11 at 17:52

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.

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Thank you very much, Noam. –  user12806 Sep 25 '11 at 16:05

I'd like to add another line of proving this (and similar problems); the proof is not yet formally complete but the method as such seems helpful for many such problems, so I show my reasoning here:

[update] After rereading my own answer I think now, that that line of attack might be more incomplete for this problem than initially thought; that the holes cannot be simply filled with one or two little more thoughts; so possibly I'll retract the whole approach later. Thanks for the upvotings anyway and apologies for distractions ... [/update]
The original answer went as below:

Let us define the following notations

a) {n,p} the exponent to which a prime p occurs in the primefactorization of some number n
b) the iverson-bracket [n:p] which evaluates to 1 if p divides n and to 0 if not

c) $\small f(a,b,n)=a^n - b^n$ , the short form for expressions where a,b are thought to be constant and n is seen as varying
d) $\small \lambda_p$ the smallest k>0 such that p divides $\small f(a,b,k)$ ,
e) $\small w_p$ the exponent, to which p occurs in $\small f(a,b,\lambda_p)$ , formally $\small w_p=\{ f(a,b,\lambda_p),p \}$

We use that definitions to rewrite the analysis of the problem in terms of the Euler-totient-theorem and the concept of the order of cyclic subgroups modulo some primes p.
Here the idea is to compare the odd primefactors in the canonical primefactorizations of the lhs and rhs in the conveniently rewritten problem
$$\small c^n-(ab)^n =^? a^{2n}+b^{2n} = {a^{4n}- b^{4n} \over a^{2n}- b^{2n} } .$$ It will be sufficient to compare the odd primefactors $\small p, q \in \text{odd primes}$ only; so we refer to possible exponents of 2 with some anonymous exponents s and t only. Then the lhs is with $\small q \in \text{odd primes}$
$$\small f(c,ab,n)=2^s \prod q^{ [n:\lambda_q] (w_q+\{n,q\})}$$ and the rhs is with some exponent t at the primefactor 2 : $$\small a^{2n}+b^{2n}={ a^{4n}-b^{4n} \over a^{2n}-b^{2n} }={f(a,b,4n)\over f(a,b,2n)}$$ and $$\small {f(a,b,4n)\over f(a,b,2n)} = 2^t { \prod p^{ [4n:\lambda_p] (w_p+\{4n,p\})} \over \prod p^{ [2n:\lambda_p] (w_p+\{ 2n , p \} ) } }$$ Here for all odd primes p we have $\small \{ 4n , p \} = \{ 2n, p \} = \{ n,p \}$, $$\small { f(a,b,4n)\over f(a,b,2n)} = 2^t \prod p^{ ([4n:\lambda_p]- [2n:\lambda_p]) (w_p+\{ n, p \} ) }$$

Conclusion: (updated)
In the rhs we get only that primefactors p, whose order divide 4n but not 2n and - having $\small n = 2^m \cdot o, o \text{ odd }$ thus must be exactly $\small 4 \cdot 2^m r$ where r is any odd divisor of n, while on the lhs we get all primes in the factorization whose order equal any divisor of n. But the sets of primes must be equal to allow a solution for the original problem.

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