Solving the Diophantine system $pqr=a^4$, $p+q+r=b^4$

Question

I am trying to find solutions of the following system of diophantine equations: $$\left\{\begin{array}{rcl}pqr&=&a^4\\p+q+r&=&b^4\end{array}\right.$$ where $a$, $b$, $p$, $q$ ans $r$ are positive integers such that $\gcd(p,q,r)$ is not divisible by $\theta^4$, $\theta>1$.

I found the following solutions $(p,q,r)$ with a computer program :

$(3\,;6\,;72)$ , $(25\,;60\,;540)$ , $(72\,;576\,;648)$ and $(162\,;448\,;686)$.

The system has infinitely many solutions : take $(p\;q\,;r)=\left(A^4\,;B^4\,;C^4\right)$, where $A^4+B^4+C^4=D^4$ and $A$, $B$, $C$ are coprime (see this article).

But can we prove that there are infinitely many solutions using more elementary ways ?

Thank you for your help !

Answer 1

Thanks to John Ca for alerting me to this question. It is equivalent to finding infinitely many solutions to the Diophantine equation $x y z (x+y+z) = 1$ in positive rational numbers $(x,y,z) = (p,q,r)/ab$. That's a special case of a problem that was already solved by Euler in 1749, though we do not know how he did it; see my lecture notes https://abel.math.harvard.edu/~elkies/euler_14t.pdf for some of the mathematical context. His formula is on page 9; taking $a=1$ we find the following parametrization (I dehomogenized by setting s=1):

$$ x = \frac{6 t^3 (t^4-2)^2} {(4 t^4 + 1) (2 t^8 + 10 t^4 - 1)}, $$ $$ y = \frac{ 3 (4 t^4 + 1)^2} {2t (t^4-2) (2 t^8 + 10 t^4 - 1)}, $$ $$ z = \frac{ 2 (2 t^8 + 10 t^4 - 1)} {3t (4 t^4 + 1)}. $$

In computer-readable format, $x,y,z$ are

6 * t^3 * (t^4-2)^2 / ((4 * t^4+1) * (2 * t^8 + 10 * t^4 - 1)),
3 * (4 * t^4+1)^2 / (2 * t * (t^4-2) * (2 * t^8 + 10 * t^4 - 1)),
2 * (2 * t^8 + 10 * t^4 - 1) / (3 * t * (4 * t^4+1))

They are all positive for $t > 2^{1/4}$; Euler gives the example $t=2$ which makes $(x,y,z) = (9408/43615, 12675/37576, 671/195)$; multiplying through by $1342^3 65^4 21$ gives the solution $$ (p,q,r) = 65^3 671^2 (1580544, 2471625, 25213496) $$ with $p+q+r = 43615^4$, $pqr = 20772769017000^4$, and $\gcd(p,q,r) = 65^3 671^2$ fourth-power free.

Some somewhat simpler parametrizations are now known, both for Euler's general problem and for this special case, but I don't think I've seen one with an explanation that is both elementary and well motivated.

Answer 2

Take,

$2p=2n^2+1-w$

$2q=2n^2+1+w$

$2r=16m^2$

Where,

$w^2=4n^4+4n^2-8m^2+1$ ---$(1)$

Eqn (1), is satisfied at, $(m,n,w)=(3,2,3)$

Hence,

$p+q+r=8m^2+2n^2+1$

For, $(m,n,w)=(3,2,3)$

$p+q+r=(3)^4$

$8pqr=(2p)(2q)(2r)$ =$(2n^2+1-w)(2n^2+1+w)(16m^2)$

For, $(m,n,w)=(3,2,3)$, we get:

$pqr=(6)^4$

$(p,q,r)=(3,6,72)$

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Answer 3

Let $q = b^4-p-r.$

We get $$-rp^2+(b^4-r)rp-a^4=0$$

Since $p$ must be rational number then discriminant must be square number.
Hence we get
$$v^2 = r^4-2b^4r^3+b^8r^2-4a^4r\tag{1}$$ Let assume $p>q>r$ then we know $r < a^{4/3}$ using $pqr=a^4$.
Hence we set $h=a^{4/3}$.
Some positive integer solutions $r$ of $(1)$ were found where $a,b,p$ are positive integers using PARI-GP.
"$hyperellratpoints(U^4-2b^4U^3+b^8U^2-4a^4U ,[h,1])$"

Though I couldn't prove the infinity of solutions, it seems there are infinitely many solutions.

$1\le a \lt 1000,\ 1\le b \lt a,\ p\gt q\gt r$

            [a , b]  [p , q , r]
            [6,    3][72, 6, 3]
            [30,   5][540, 60, 25]
            [42,   8][3087, 1008, 1]
            [72,   6][648, 576, 72]
            [84,   6][686, 448, 162]
            [120,  7][1600, 576, 225]
            [168, 14][37632, 756, 28]
            [180,  9][5625, 648, 288]
            [204, 17][83232, 153, 136]
            [252,  9][3528, 2592, 441]
            [252, 11][9072, 5488, 81]
            [264, 37][1874048, 81, 32]
            [294, 18][100842, 4116, 18]
            [378, 12][18522, 1458, 756]
            [420, 10][4900, 2940, 2160]
            [462, 11][8712, 4851, 1078]
            [468, 26][454896, 2028, 52]
            [504, 17][76832, 6561, 128]
            [510, 13][18360, 9826, 375]
            [624, 21][165888, 28561, 32]
            [630, 14][34020, 2646, 1750]
            [630, 26][455625, 1008, 343]
            [700, 15][30625, 19600, 400]
            [702, 21][118638, 75816, 27]
            [714, 17][77112, 5831, 578]
            [780, 20][131820, 28080, 100]
            [840, 13][14400, 11025, 3136]
            [924, 19][115248, 14641, 432]
            [990, 15][27225, 21780, 1620]

Answer 4

We have the below Identity:

$(b^2+c^2-a^2)^2+(2ab)^2+(2ac)^2=(b^2+c^2+a^2)^2$

Let:

$p=(b^2+c^2-a^2)^2$

$q=(2ab)^2$

$r=(2ac)^2$

$p+q+r=(a^2+b^2+c^2)^2$

To make (RHS) a fourth power we take:

$b^2+c^2=8a^2$ ---$(1)$

Hence,

$p+q+r=(9a^2)^2=(3a)^4$

$(pqr)=((b^2+c^2-a^2)(2ab)(2ac))^2$ =$(2a)^4(bc(b^2+c^2-a^2))^2$

Since, $(b^2+c^2=8a^2)$ we have $(b^2+c^2-a^2)=(7a^2)$ and to make, $(bc(b^2+c^2-a^2))$, a square we take,

$b=7c$ ---$(2)$

Hence,

$pqr=(2a)^4(7ac)^4=(14a^2c)^4$

Now eqn (1) & (2) above is satisfied by $(a,b,c)=(5,14,2)$

Hence:

$(p+q+r)=(15)^4$

$(p,q,r)=(700)^4$

And,

$(p,q,r)=(30625,19600,400)$

Answer 5

Here is a possible "minimal" solution, based on the arithmetic of elliptic curves. So we are trying to realize the second equation constraining the sum $p+q+r$ in the question with three terms of the shape $p=s^2t^2$, $q=t^2u^2$, $r=u^2s^2$, and we are changing the constant-like variable $b$ into a variable-like letter $v$. So we would like to solve and get infinitely many solutions for $$ \tag{$*$} \color{blue}{u^2}(s^2+t^2) + s^2t^2 = \color{blue}{v^4}\ . $$ Even when we succeed this for a particular value of $(s,t)$, we are done. (The product $pqr$ is already a fourth power.)

(Note: There is already an accepted solution using the splitting $s^3tu+st^3u+stu^3=v^4$ instead, it provides an explicit one-parameter family. Having the family, the check is more or less elementary, but finding it runs into an arithmetic search over function fields in one variable. This is in nature comparable with the effort of getting a parametrization for $A^4+B^4+C^4=D^4$, considered to be involved by the OP.)

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For the special tuple $\color{maroon}{(s,t) = (1,16)}$ the corresponding affine curve in $(*)$ is $$ C\ :\qquad \color{maroon}{257}\;\color{blue}{u^2} + \color{maroon}{256} = \color{blue}{v^4} \ , $$ so it is enough to show there are infinitely many $\Bbb Q$-rational points on this particular curve $C$. For such a rational triple, after clearing denominators in a minimal manner (introducing a fourth power in the sum) we obtain an integer solution in the spirit of the OP. Regarding the solutions / the points in $C(\Bbb Q)$, we have the following result:

Lemma: Let $E$ be the elliptic curve of rank two $$ E\ :\qquad \eta^2 = \xi^3 + 67634176\;\xi\ . \qquad\text{(Here $67634176=4\cdot 256\cdot 257^2$.)} $$ with generators $$ P = (1568,\ 331520)\ ,\ Q = (3281633/169,\ 6455959505/2197) \ . $$ Then for a generic $(\xi,\eta)$ in the infinite group $E(\Bbb Q)$ the following point $(u,v)$ is in $C(\Bbb Q)$: $$ u=256\cdot\frac \eta{(\xi-d)^2}\ ,\ v=4\cdot\frac {\xi+d}{\xi-d}\ . \qquad\text{Here $d:=8224=2^5\cdot 257$.)} $$

Proof: Human computation, inserted here for the complete proof: $$ \begin{aligned} (257\; u^2 + 256)(\xi-d)^4 &= 257\cdot 256^2\eta^2 + 256(\xi-d)^4 \\ &= 257\cdot 256^2\cdot (\xi^3 + 4\cdot 256\cdot 257^2\;\xi) + 256(\xi-d)^4 \\ &= 256\cdot 8d\cdot (\xi^3 + d^2\;\xi) + 256(\xi-d)^4 \\ &= 256\cdot\Big[\ (\xi-d)^4 + 8d\;\xi^3 + 8d^3\;\xi) + 256\ \Big] \\ &= 4^4\cdot(\xi+d)^4 \ .\qquad\text{Now divide by $(\xi-d)^4$ to get:} \\ 257\; u^2 + 256 &=v^4\ . \end{aligned} $$ $\square$

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The sage code for the computer check/proof is:

var('X,Y');
d = 8224
u, v = 256*Y/(X - d)^2, 4*(X + d)/(X - d)
factor(257*u^2 + 256 - v^4)

And the factorization reveals the needed factor, the defining polynomial of $E$:

-16842752*(X^3 - Y^2 + 67634176*X)/(X - 8224)^4

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Note: This solution could be considered elementary, together with the parallel geometric argument of producing infinitely many points $jP+kQ\in E(\Bbb Q)$. We obtain at any rate infinitely many different points and need no height complexity argument, since else we would have infinitely many torsion points, but the torsion is easily controlled.

We are done, but i will give some more details on how this solution was found. First of all, start with the curve $C$. Sage gives the Weierstraß form for it, and a map in the wrong direction. In a dialog with the machine in the ipython-sage interpreter:

sage: R.<u,v> = PolynomialRing(QQ)
sage: E = EllipticCurve(QQ, WeierstrassForm(257*u^2 + 256 - v^4))
sage: E
Elliptic Curve defined by y^2 = x^3 + 67634176*x over Rational Field

sage: E.rank()
2
sage: E.gens()
[(1568 : 331520 : 1), (3281633/169 : 6455959505/2197 : 1)]

sage: X, Y, Z = WeierstrassForm(257*u^2 + 256 - v^4, transformation=True)

sage: X, Y, Z
(67634176*v^2, 17381983232*v^5 + 4449787707392*v, 257*u)
sage: factor(-Y^2 + X^3 + 67634176*X*Z^4)
(-302133341077529165824) * v^2 * (v^4 - 257*u^2 - 256) * (v^4 + 257*u^2 - 256)

And the needed factor $(v^4 - 257u^2 - 256)$ is indeed in there. Moving the story in a mathematical frame, the map in the wrong direction joining $C$ and the elliptic curve $E$ is $(u,v)\to (X/Z^2,\ Y/Z^3)$ with the above notations, so it is: $$ \begin{aligned} \xi &=\bar\xi(u,v):= 1024\cdot\frac {v^2}{u^2}\ ,\\ \eta &=\bar\eta(u,v):=1024\cdot\frac vu\cdot\frac{v^4+256}{u^2}\ . \\[2mm] \text{Check:} & \\ \xi^3+ 67634176\;\xi &=\xi(\xi^2+ 4\cdot 256\cdot 257^2) \\ &=\frac 1{u^6}\cdot 1024\; v^2\;(1024^2\;v^4 + 4\cdot 256\cdot 257^2\; u^4) \\ &=\frac 1{u^6}\cdot 1024\; v^2\;(1024^2\;(256+257 u^2) + 4\cdot 256\cdot 257^2\; u^4) \\ &=1024^2\cdot\frac{v^2}{u^2} \cdot\frac {(2\cdot 256+257u^2)^2}{u^4} \\ &=\eta^2\ . \end{aligned} $$

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Now the question is if this is a birational map. No it is not. Not every point $R$ in $E(\Bbb Q)$ has a preimage by the above map, but each point of the shape $2R$ has. This is enough to get infinitely many points in $C(\Bbb Q)$. We ask for the formula for the map $R\to 2R$, and (can) solve for each $(\xi,\eta)\in E(\Bbb Q)$ the corresponding system $$ \left\{ \begin{aligned} \xi_2 &= \bar\xi(u,v)\ ,\\ \eta_2 &= \bar\eta(u,v)\ , \end{aligned} \right. \qquad\text{ where }(\xi_2,\eta_2):=2\odot(\xi,\eta)\text{ on }E(\Bbb Q)\ , $$ as a system in $(u,v)$. The solution was published in the short Lemma above.

All the story so far went well after the knowledge of a good starting tuple $(s,t)=(1,16)$. It turns out that some relatively few, but not so rare further tuples can be found to work with in the spirit of the above solution. In all other "simple"/small cases the $(s,t)\ne(1,16)$ leads to a related elliptic curve $E$ with a similar map in wrong direction, with rank zero or one. Rank zero is of no use. When the rank is one, with a generator denoted by $P$, the points $R\in E(\Bbb Q)$ that are in the image of the map turn out to be those of the shape $kP$ with an odd $k$. I could not find in the given time a pair $(s,t)$ leading to an $E$ of rank three or higher to see what happens with the image in this case.

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For illustration, here are some points $(u,v)$ found on $C$ by the above receipt. $$ \left(\ \pm\frac{1295}{676}\ ,\ \pm\frac{153}{26}\ \right)\ ,\ \left(\ \pm\frac{7454415}{327184}\ ,\ \pm\frac{10937}{572}\ \right)\ ,\ \left(\ \pm\frac{325295360}{54184321}\ ,\ \pm\frac{72708}{7361}\ \right)\ , \\ \left(\ \pm\frac{10011924802896945}{15876092736135424}\ ,\ \pm\frac{548157497}{126000368}\ \right)\ ,\ \left(\ \pm\frac{40983543870880000}{32507836498694209}\ ,\ \pm\frac{915409284}{180299297}\ \right)\ , \left(\ \pm\frac{876949723117278720}{5683360042891969}\ ,\ \pm\frac{3749515652}{75388063}\ \right)\ , \\ \left(\ \pm\frac{323043246288601617984248527921099520}{444899698827746005222326219257575489}\ ,\ \pm\frac{2966970397567963524}{667008020062537183}\ \right)\ ,\ \\ \left(\ \pm\frac{27405441078951726763829075599483610665766263824876237447371487778467840}{16101404754550592670709225929792188672621384173959898871314853205067969}\ ,\ \pm\frac{713655883428831684864440220067318148}{126891310792152323288347809785654687}\ \right)\ ,\ \dots $$

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Some other good pairs $(s,t)$ with a corresponding elliptic curve of rank alt least two are...

R.<u, v> = QQ[]
for s, t in cartesian_product([[1..40], [1..40]]):
    if s > t or gcd(s, t) > 1:
        continue
    quartic = u^2*(s^2 + t^2) + s^2*t^2 - v^4
    E = EllipticCurve(QQ, WeierstrassForm(quartic))
    try:
        rk = E.rank(only_use_mwrank=False)
        if rk < 2:
            continue
    except:
        continue
    print(f'{s} & {t} & {rk}\\\\\\hline')

Results: $$ \begin{array}{|r|r|r|} \hline s & t & \operatorname{rank}E\\\hline \color{maroon}{1} & \color{maroon}{16} & \color{maroon}{2}\\\hline 1 & 33 & 2\\\hline 1 & 34 & 2\\\hline 3 & 11 & 2\\\hline 3 & 19 & 2\\\hline 3 & 23 & 2\\\hline 4 & 17 & 2\\\hline 4 & 35 & 2\\\hline 5 & 8 & 2\\\hline 5 & 21 & 2\\\hline 6 & 7 & 2\\\hline 7 & 15 & 2\\\hline 8 & 13 & 2\\\hline 9 & 17 & 2\\\hline 10 & 21 & 2\\\hline 10 & 37 & 2\\\hline 11 & 15 & 2\\\hline 11 & 24 & 2\\\hline 11 & 27 & 2\\\hline 11 & 35 & 2\\\hline 13 & 21 & 2\\\hline 13 & 32 & 2\\\hline 15 & 23 & 2\\\hline 16 & 35 & 2\\\hline 17 & 33 & 2\\\hline 17 & 36 & 2\\\hline 19 & 27 & 2\\\hline 19 & 33 & 3\\\hline 20 & 29 & 2\\\hline 20 & 39 & 2\\\hline 21 & 25 & 2\\\hline 22 & 39 & 2\\\hline 23 & 35 & 2\\\hline 23 & 39 & 2\\\hline 25 & 34 & 2\\\hline 26 & 29 & 2\\\hline 27 & 34 & 3\\\hline 27 & 40 & 3\\\hline 28 & 29 & 2\\\hline \end{array} $$ and so on. Compared to the short above list, there are very, very many tuples $(s,t)$ with a value $1$ of the rank of the corresponding elliptic curve. But there is a rare rank $3$ showing up in the given computing time during this brute force search in a modest domain with no arithmetical further insight. (During the last three nights the stated problem became secondary for me, i was highly attracted by the arithmetic inside this special family $(*)$ of curves.)