# For what integer $n$ are there infinitely many $-a+nb+c = -d+ne+f$ where $a^6+b^6+c^6 = d^6+e^6+f^6$?

(Much revised for clarity.) I was considering the system of equations,

$$-a+nb+c = -d+ne+f\tag1$$ $$a+b+c = d+e+f\tag2$$ $$a^2+b^2+c^2 = d^2+e^2+f^2\tag3$$ $$a^6+b^6+c^6 = d^6+e^6+f^6\tag4$$

Question 1. Is it true that, for a fixed integer $n$, if the system has an infinite number of co-prime integer solutions, then $n$ is a multiple of $3$?

Method: Eqs $(2)$ and $(3)$ can easily be given a complete solution. Incorporating $(1)$, I got,

$$(-2 p + \alpha q -\beta u)^k + (\beta p - 2 u)^k+(\beta q + \alpha u)^k =\\ (-2 p + \alpha q +\beta u)^k + (\beta p + 2 u)^k+(\beta q -\alpha u)^k\tag5$$

where $\alpha = n+1,\;\beta = n-1$. It is also true for $k=6$ if there is $p,q,n$ such that,

$$Poly_1:= (-3+n)(5-2n+n^2)p + 4n(1+n^2)q$$

$$Poly_2:= (-3+n)(5-2n+n^2)p^3 + 2(5+11n-5n^2+n^3)p^2q - (5+7n+15n^2-3n^3)pq^2 + 4n(1+n^2)q^3$$

and,

$$\color{red}{-}Poly_1 Poly_2 = \text{square}\tag6$$

A trivial solution is $q = \frac{(3-n)p}{2n}$ which yields,

$$\color{red}{-}Poly_1 Poly_2 = \frac{(-9+n^2)^2(-1+n)^4p^4}{4n^2}$$

Example: Let $n=12$, then,

$$((-2 p + 13 q - 11 u)^k+(11 p - 2 u)^k+(11 q + 13 u)^k =\\(-2 p + 13 q + 11 u)^k+( 11 p + 2 u)^k+( 11 q - 13 u)^k\tag7$$

which is already true for $k=1,2$. But it is also for $k=6$ if,

$$225 p^3 + 458 p^2 q + 587 p q^2 + 1392 q^3 = -3 (75 p + 464 q) u^2\tag8$$

An initial point is $p,q = -104,17.$ Hence $(8)$ can be easily turned into an elliptic curve, so there is an infinite number of integer solutions to $(7)$.

Question 2. For what other positive integer $n$ below a bound can we find solutions to non-zero $(6)$ or with $(3-n)p-2nq \neq 0$? (The constraint is to prevent trivial solutions. I have found $n=12, 15, 21, 30, 33, 135$ but I am not sure if this is exhaustive for $n<150$.)

• Incidentally, the smallest solution to $a^6+b^6+c^6 = d^6+e^6+f^6$, namely, $$(-23)^6+ 10^6+ 15^6 = 3^6+ 19^6+ (-22)^6$$ belongs to another family that is also true as $$3a+b+c = d+e+3f$$ – Tito Piezas III Nov 5 '15 at 5:05
• I trust you're familiar with Bremner, A geometric approach to equal sums of sixth powers, Proc. London Math. Soc. (3) 43 (1981), no. 3, 544–581, MR0635569 (83g:14018), and Delorme, On the Diophantine equation $x^6_1+x^6_2+x^6_3=y^6_1+y^6_2+y^6_3$, Math. Comp. 59 (1992), no. 200, 703–715, MR1134725 (93a:11023) and Choudhry, On equal sums of sixth powers, Indian J. Pure Appl. Math. 25 (1994), no. 8, 837–841, MR1293455 (95k:11036) etc., etc. – Gerry Myerson Nov 5 '15 at 22:43
• @GerryMyerson: Of course, but Bremner's and Delorme's paper deals with the constraint that, $$a^2 +ad−d^2 =−(b^2 −be−e^2 )\\b^2 +be−e^2 =−(c^2 −cf−f^2 )\\c^2 +cf−f^2 =−(a^2 −ad−d^2 )\tag1$$ while Choudhry's has, $$a+b+c=d+e+f\tag2$$ This post does not obey $(1)$, but instead focuses on the constraint, $$a+nb+c=d+ne+f\tag3$$ and for what integer $n$ are permissible. – Tito Piezas III Nov 6 '15 at 1:56
• @GerryMyerson: I've changed the title to better reflect my intent. :) – Tito Piezas III Nov 6 '15 at 4:24
• Probably need to reformulate the problem as the solution of the system of equations $(1)$. – individ Nov 6 '15 at 10:49

I am sorry to say that the answer to Question 1 is NO.

As you state, $q=(3-n)p/(2n)$ gives a solution to the problem. Thus the quartic must be birationally equivalent to an elliptic curve.

Using standard methods, it is possible to show that this elliptic curve is of the form \begin{equation*} G^2=H^3+25( h_1 H + h_2)^2 \end{equation*} where \begin{equation*} h_1=(n^2+3) \hspace{1cm} h_2=4(n^2+1)(n^2+2n+5)(n^2-2n+5) \end{equation*} with the reverse transformation \begin{equation*} \frac{p}{q}=\frac{2n\big(G+(n^2+11)H+5h_2\big)}{(3-n)\big(G+(3n^2+4n+5)H+5h_2\big)} \end{equation*}

The elliptic curves have torsion points when $H=0$, and the torsion subgroup seems to be $\mathbb{Z}3$.

Applying the Birch and swinnerton-Dyer conjecture to these curves, the first positive rank is at $n=12$ as you state, but the second is at $n=13$, where $H=-608000/9$ gives a rational point. Substituting backwards we arrive at the solution to the initial problem for $n=13$: $a=-1181$, $b=-6033$, $c=6302$, $d=6805$, $e=-4702$ and $f=-3015$.

I computed height estimates for $n$ up to $100$. There seems no obvious pattern and some of the heights are very large, suggesting a solution but one with enormous numbers. Since the elliptic curves do not have a point of order $2$, finding these solutions will not be easy.

• Thanks! With $n=13$, I get, $$-a+13b+c = -d+13e+f$$ where $$(-p + 7 q - 6 u)^k+(6 p - u)^k+( 6 q + 7 u)^k = \\(-p + 7 q + 6 u,)^k+(6 p + u)^k+( 6 q - 7 u)^k$$ and already true for $k=1,2$. It is also for $k = 6$ if $$37 p^3 + 75 p^2 q + 99 p q^2 + 221 q^3 = -(37 p + 221 q) u^2$$ From your transformations, I get initial points $p/q = -565/173$ and $p/q = -1049/193$. No wonder I was not able to find this $n$ since I only limited the search to $p < \pm500$ and $q<200$. – Tito Piezas III Nov 9 '15 at 13:18
• By the way, if you want, can you add the list of $n\leq100$ you found that are solvable? It will be a useful piece of information to your answer. – Tito Piezas III Nov 10 '15 at 2:55

As asked by OP here are values of $n$ where the BSD Conjecture predicts a strictly positive rank for the elliptic curve

12, 13, 15, 16, 17, 18, 21, 23, 24, 25, 26, 27, 30, 32, 33, 35, 36, 38, 40, 41, 43, 44, 47, 49, 52, 53, 54, 56, 57, 64, 65, 66, 67, 69, 73, 75, 76, 78, 79, 80, 81, 85, 86, 88, 90, 91, 93, 94, 95, 97.