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(Note: See also the $a^4+b^4+c^4 = 1$ version in this old MSE post.)

The equation discussed in a paper by Jacobi and Madden,

$$a^4+b^4+c^4+d^4 = (a+b+c+d)^4 = z^4\tag1$$

or equivalently,

$$(p-2q + r)^4 + (p-2q - r)^4 + (q + s)^4 + (q - s)^4 = (2p - 2q)^4\tag2$$

After some algebra, $(2)$ can be solved as an intersection of two rather simple quadric surfaces,

$$(m^2-7) p^2 + 24pq-24q^2= (m^2+1) r^2\tag3$$

$$8mp^2-24mpq - 3(m^2 - 8m + 1) q^2 = (m^2 + 1) s^2\tag4$$

for some constant $m$. Given a known solution to $(1)$, $m$ can be recovered as,

$$m =\frac{3q^2+s^2}{p^2-r^2}\tag5$$

There are only two (?) known primitive solutions to $(1)$ with $z<220000$.

I. Solution 1:

$$(-2634)^4+5400^4+1770^4+955^4 = (-2634+5400+1770+955)^4=5491^4$$

From these $a,b,c,d$, after permutation one can get six distinct values for $m$,

$$m_k =\frac{511}{450}, \, \frac{31^2}{61}, \, \frac{1423}{1098}, \, \frac{2521}{325}, \, \frac{1651}{126},\,\frac{1777}{1525}\tag6$$

which are, in fact, (courtesy of a comment by Jeremy Rouse),

$$m_k =\frac{n_1}{d_1},\,\frac{n_1+d_1}{n_1-d_1},\,\frac{n_3}{d_3},\,\frac{n_3+d_3}{n_3-d_3},\,\frac{n_5}{d_5},\,\frac{n_5+d_5}{n_5-d_5}$$

For example, using $m_2 =\frac{31^2}{61}$, and applying it to $(3),(4)$, we get,

$$448737 p^2 + 44652 p q - 44652 q^2 = 463621 r^2$$

$$234484 p^2 - 703452 p q - 687411q^2 = 463621 s^2$$

which has initial rational $p,q = -1906,\,\frac{1679}{2}$. Using an elliptic curve, we can then get an infinite more.

II. Solution 2:

$$(-31764)^4+ 27385^4+ 48150^4+ 7590^4 = (-31764+27385+ 48150+7590)^4 = 51361^4$$

From these, we get,

$$m_k =\frac{193}{18},\, \frac{211}{175},\, \frac{619}{450},\,\frac{1069}{169},\, \frac{1141}{666},\, \frac{1807}{475}\tag7$$

For example, using $m_1 = \frac{193}{18}$, with initial $p,\,q = \frac{27187}{2}, -12087$.

Question:

  1. Without knowing solutions 1 and 2 in advance, is there a systematic way to search for appropriate $m$ such that $(3),\,(4)$ has rational solutions? What other $m$ is there of small height not in the list of twelve above?
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  • $\begingroup$ I know I am revisiting an old question. As I don't have access to The American Mathematical Monthly I don't have direct access to the original paper of Jacobi and Madden. Is there anywhere else where one might read about their method in more detail? $\endgroup$ – Jesper Petersen Nov 19 '15 at 15:42
  • $\begingroup$ @JesperPetersen Surely the library of a nearby university will have access? Anyway, while it will be informative to read the original paper, you won't find $(3)$ and $(4)$ in their paper since I derived it from ideas they outlined. In fact, if I recall correctly, they used only one elliptic curve. (Though that is enough to prove the equation in question has an infinite number of primitive solutions.) $\endgroup$ – Tito Piezas III Nov 19 '15 at 16:02
  • $\begingroup$ Thanks for the suggestion. I ordered a copy of the paper from a local university library. I was hoping for an easily accessible online copy, but this will do, and will hopefully be both interesting and informative to read in more detail. $\endgroup$ – Jesper Petersen Nov 19 '15 at 18:09
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Given a rational number $m$, let $C_{m}$ be the intersection of two quadrics given by (3) and (4) in the original question. The way to search for points is to test the curve $C_{m}$ to see if it has local points. I did this for all $m$ with height $\leq 193$ (since I knew I'd find points on $C_{193/18}$). There are only $18$ rational numbers for which $C_{m}$ has points in $\mathbb{R}$, and $\mathbb{Q}_{p}$ for $p = 2$ an all primes of bad reduction of the Jacobian of $C_{m}$. (I did this computation in Magma, and it was complicated by the fact that Magma currently has a bug in its IsLocallySolvable command.)

(The $18$ rationals of height less than or equal to $193$ for which $C_{m}$ has local points are $157/150$, $163/150$, $151/126$, $181/150$, $121/96$, $103/78$, $97/72$, $79/54$, $37/25$, $67/42$, $49/24$, $73/25$, $109/25$, $31/6$, $133/25$, $169/25$, $181/25$, and $193/18$.)

Searching for points on these $C_{m}$ of height up to $10^{8}$ reveals points for $m = 157/150$, $m = 37/25$, $m = 31/6$ and the point on $m = 193/18$ that we already knew about. New solutions to (1) that arise from these points are \begin{align*} a &= 841263, b = -792940, c = 3852350, d = -44410\\ a &= 35847220, b = -34122866, c = 53902630, d = 2542025\\ a &= 39913670, b = -23859495, c = 15187700, d = 10116014,\\ \end{align*} which come from $m = 157/150$, $m = 37/25$ and $m = 31/6$, respectively. (I actually found two points on the $m = 31/6$ curve. The other point gives a solution to (1) which is a permutation of the $a = 35847220$ solution.)

It may be possible to prove that $C_{m}$ has no points on it for other choices of $m$ using the fact that $C_{m}$ is a four-cover of its Jacobian and using other techniques (the Cassels-Tate pairing, 3-descents, $L$-functions) to show that $C_{m}$ represents an element of Sha.

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  • $\begingroup$ Great! Also, because of your comment about $m_1=31/6$ and $m_2 =37/25$ being related by permutation, I notice (fortunately they are small) that $m_2 = (31+6)/(31-6)$. I checked the first six $m$ given by $(6)$ and, arranged properly, they were in fact, $$m_k =\frac{n_1}{d_1},\,\frac{n_1+d_1}{n_1-d_1},\,\frac{n_3}{d_3},\,\frac{n_3+d_3}{n_3-d_3},\,\frac{n_5}{d_5},\,\frac{n_5+d_5}{n_5-d_5}$$ Persumably this happens in general. Do you know why? $\endgroup$ – Tito Piezas III Dec 23 '14 at 20:39
  • $\begingroup$ Yes, this does happen in general. Viewing $m$ as a function of $a$, $b$, $c$ and $d$ we can see the effect on $m$ of permuting $a$, $b$, $c$ and $d$. There's one permutation of the variables that will send $m$ to $(m+1)/(m-1)$ (under the assumption that $(a+b+c+d)^4 = a^4 + b^4 + c^4 + d^4$). It turns out that the Jacobians of $C_{m}$ and $C_{(m+1)/(m-1)}$ are isomorphic. For a given solution $(a,b,c,d)$ we get three isomorphism classes of Jacobians, so there's probably no relation between $m_{1}$, $m_{3}$ and $m_{5}$. $\endgroup$ – Jeremy Rouse Dec 23 '14 at 23:46
  • $\begingroup$ I find that if we define, $$\begin{align}x_1 &= m_1+m_3+m_5-1\\ x_2 &= m_1m_3+m_1m_5+m_3m_5\\ x_3 &= m_1m_3m_5\end{align}$$ then, $$x_1^2-x_1x_2+x_2^2-3x_1x_3 = - 3$$ where either $m_5 = m_5',$ or $m_5 = \frac{m_5'+1}{m_5'-1}$. $\endgroup$ – Tito Piezas III Jul 21 '15 at 6:45
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Seiji Tomita found a fifth elliptic curve $E$ as $m = \frac{19^2}{61}$ with an explicit solution. In summary, there are now seven "small" primitive solutions to,

$$a^4+b^4+c^4+d^4 = (a+b+c+d)^4 = z^4$$

or equivalently,

$$(p-2q + r)^4 + (p-2q - r)^4 + (q + s)^4 + (q - s)^4 = (2p - 2q)^4$$

with $z$ less than 60 million:

$$\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline E&m&a&b&c&d&z&\text{by}&\text{year}\\ 1&\color{blue}{\frac{31^2}{61}}&5400& 1770& -2634& 955& 5491& \text{Brudno}&1964\\ 2&\frac{211}{175}&-31764& 7590& 27385& 48150& 51361& \text{Wroblewski}&2005\\ 3&\color{blue}{\frac{19^2}{61}}&1984340& -107110& 1229559& -1022230& 2084559& \text{Tomita}&2015\\ 3&\color{blue}{\frac{19^2}{61}}&3597130& -1953890& 561760& 1493309& 3698309& \text{Tomita}&2015\\ 4&\frac{157}{150}&841263& -792940& -44410& 3852350& 3856263& \text{Rouse}&2014\\ 5&\frac{31}{6}&39913670& -23859495& 15187700& 10116014& 41357889& \text{Rouse}&2014 \\ 5&\frac{31}{6}&53902630& 2542025& 35847220& -34122866& 58169009& \text{Rouse}&2014\\ \hline \end{array}$$

The variable $m$ can be recovered as,

$$m =\frac{3q^2+s^2}{p^2-r^2}$$

(I have no idea if $\displaystyle\frac{19^2}{61}$ and $\displaystyle\frac{31^2}{61}$ is just coincidence.) Incidentally, as pointed out by Rouse, if $m$ is solvable, then so is $m'=\frac{m+1}{m-1}$. Hence, Tomita's $m = \frac{211}{150}$ in his website yields $m' = \frac{19^2}{61}$ cited here.

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It is not an answer to the question but rather a short note about how one can find algebraic relations between the quantities $m_1, m_2, m_3, m_4, m_5, m_6$. First, let us verify directly that there is only one relation between $m_1$ and $m_2$ (we use Magma Calculator):

> A4<a,b,c,d>:=AffineSpace(Rationals(),4);
> S:=Scheme(A4,a^4+b^4+c^4+d^4-(a+b+c+d)^4);
> A<m1,m2>:=AffineSpace(Rationals(),2);
> f:=map<S->A|[(c^2+c*d+d^2)/((a+c+d)*(b+c+d)),
>    (a^2+a*b+b^2)/((a+c+b)*(b+a+d))]>; 
> Image(f);
Scheme over Rational Field defined by
m1*m2 - m1 - m2 - 1

As you can see, the idea was to construct an appropriate map from the affine variety $$S:=\{ (a,b,c,d) \in \mathbb Q^4 \ | \ a^4 +b^4 +c^4 +d^4=( a+b+c+d)^4 \} $$ to the two-dimensional affine space. Using the same approach one get all relations between $m_1, m_3, m_5$:

> A<m1,m3,m5>:=AffineSpace(Rationals(),3);
> f:=map<S->A|[(c^2+c*d+d^2)/((a+c+d)*(b+c+d)),
> (a^2+a*d+d^2)/((a+c+d)*(b+d+a)),
> (b^2+b*d+d^2)/((a+b+d)*(b+d+c))]>; 
> Image(f); 
Scheme over Rational Field defined by
m1^2*m3^2*m5^2 - 2*m1^2*m3^2*m5 + m1^2*m3^2 - 2*m1^2*m3*m5^2 + 2*m1^2*m3*m5 +
    m1^2*m5^2 + 3*m1^2 - 2*m1*m3^2*m5^2 + 2*m1*m3^2*m5 + 2*m1*m3*m5^2 -
    8*m1*m3*m5 + 2*m1*m3 + 2*m1*m5 - 6*m1 + m3^2*m5^2 + 3*m3^2 + 2*m3*m5 - 6*m3
    + 3*m5^2 - 6*m5 + 7
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