$x^4+ry^4=z^4$: find primitive Diophantine solutions with prime $r$.

Background. I tried to find an answer to this question Diophantine equation $x^4+5y^4=z^4$ that gives a prime primitive solution $(x,y,z,r)=(1,2,3,5)$, but I was unable to make any progress, so I looked at the more general $x^4+ry^4=z^4$.

My efforts. Clearly $r=0$ is trivial and the case $r<0$ is easily bypassed by swapping $x,z$ values. I can see that $r$ cannot be the fourth power of an integer, due to Fermat’s Last Theorem. I’ve noticed that without the restriction that $r$ is prime, primitive solutions are overabundant.

1. I rearranged the equation as $r=(z^4-x^4)/y^4$, then $$r=(z^4-x^4)/y^4=(z-x)(z+x)(z^2+x^2)/y^4$$ Set $f_1=(z-x)$, $f_2=(z+x)$, and $f_3=(z^2+x^2)$ then $r=(f_1f_2f_3/y^4)$

  1. Starting with the prime primitive solutions $(x,y,z,r)=(1,2,3,5)$, I considered that, maybe, $(f_1,f_2,f_3)$ are given by $(y,y^2,ry)$ This gives $$(x,y,z,r)=(\frac{(y^2-y)}2,y,\frac{y^2+y}2,\frac{y(y^2+1)}2)$$

but clearly, $r$ is composite except for $y=2$ and the trivial $y=1$ $$(x,y,z,r)=(1,2,3,5)$$

  1. I tried setting $f_3$ to a fourth power, and found a parametric solution, $$(x,y,z,r)=(Abs(-a^4+6a^2b^2-b^4), a^2+b^2, 4ab(a^2-b^2),Abs((a^4+4a^3b-6a^2b^2-4ab^3+b^4)*(a^4-4a^3b-6a^2b^2+4ab^3+b^4))$$ Unsurprisingly, I found just the single solution with $r$ prime,


  1. When $f_1=b^4$ and $f2=a^4$ we have, $z-x=b^4$ and $z+x=a^4$, then,


Clearly, both $a$ and $b$ must be odd, and $gcd(a,b)=1$ are necessary, but it’s inevitable that composite $r$ will be also generated. Here are the smallest prime $r$, I’ve found of this type, shown as $(a,b,x,y,z,r)$

$$(5,3,272,15,353,198593)$$ $$(9,1,3280,9,3281,21523361)$$ $$(11,3,7280,33,7361,107182721)$$ $$(13,1,14280,13,14281,407865361)$$ $$(13,9,11000,117,17561,429388721)$$ $$(17,3,41720,51,41801,3487882001)$$ $$(19,17,23400,323,106921,11979660241)$$ $$(23,3,139880,69,139961,39155495921)$$ $$(25,19,130152,475,260473,84785726833)$$ $$(27,7,264520,189,266921,141217650641)$$ $$(27,11,258400,297,273041,141321947681)$$ $$(29,5,353328,145,353953,250123401793)$$ $$(29,17,311880,493,395401,253611085201)$$ $$(29,23,213720,667,493561,289278699121)$$ $$(29,27,87920,783,619361,391337974721)$$ $$(31,5,461448,155,462073,426445714033)$$ $$(31,7,460560,217,462961,426448401121)$$ $$(31,11,454440,341,469081,426552698161)$$ $$(31,27,196040,837,727481,567660286961)$$ $$(31,29,108120,899,815401,676568725201)$$ $$(33,1,592960,33,592961,703204309121)$$ $$(37,3,937040,111,937121,1756239730241)$$ $$(37,27,671360,999,1202801,1897454495201)$$ $$(39,31,694960,1209,1618481,3102450148961)$$ $$(41,23,1272960,943,1552801,4031618107201)$$ $$(43,1,1709400,43,1709401,5844100138801)$$ $$(43,9,1706120,387,1712681,5844121662161)$$ $$(43,15,1684088,645,1734713,5845381584113)$$ $$(43,17,1667640,731,1751161,5847588017521)$$ $$(45,37,1113232,1665,2987393,10163802422273)$$

Updated 15 Jan 2017.

My questions.

  1. Apart from the solutions I’ve found, are there any more? I’m looking for numerical solutions and/or formulae that produce candidates for prime $r$.
  2. Does the equation $x^4+ry^4=z^4$ have a name?
  3. Are there any relevant papers etc available online.
  4. Can anyone prove there are no further solutions?
  5. Any helpful ideas as to further progress, please?
  • 1
    $\begingroup$ Note that your #3 shows this equation’s intimate connection with Ljunngren’s equation $X^2+1=2Y^4$. $\endgroup$ – Kieren MacMillan Dec 7 '17 at 19:25
  • $\begingroup$ Thank you for your interest in this question. I guess both solutions containing $13$ and $239$ is far more than coincidence, and would be fascinated to know more. $\endgroup$ – Old Peter Dec 8 '17 at 19:17
  • 1
    $\begingroup$ Not just $13$ and $239$: Ljunggren's triangle $(119,120,169)$ is the only integer right triangle with a square hypotenuse, where $169=13^2$. And $(3,2)$ is the largest (and only non-trivial) solution to the Thue equation which is derived from Ljunngren's equation. In other words, all six of your $(a,b,x,y,z,r)$ have a corresponding element in Ljunngren's problem. $\endgroup$ – Kieren MacMillan Dec 9 '17 at 6:00
  • $\begingroup$ Maybe I can interest you in this post on the next step $x^5+ny^5= z^5$? $\endgroup$ – Tito Piezas III Jan 13 '18 at 17:15

I. Let $r =n$. It may be productive to solve the simpler equation $$e^2 + n f^2 = g^2$$

The complete primitive solution has long known to be $$(p^2-nq^2)^2+n(2pq)^2 =(p^2+nq^2)^2$$

So your problem reduces to solving the non-primitive system, $$p^2-nq^2 = \pm k\, t_1^2\\p^2+nq^2 = k\,t_2^2\\2pq = k\, t_3^2$$

The equation, $$(p^2-nq^2)(p^2+nq^2) =p^4-n^2q^4= t^2$$ is well-studied, and deals with congruent numbers $n$. Thus the equation, $$x^4 + n y^4 = z^4$$

is a subset of the congruent number problem but with a third constraint. Example, let $u=15^2,\,v=1$, then,

$$u^2 - 198593v^2 = -2\times272^2\\ u^2 + 198593v^2 = 2\times353^2\\ 2uv = 2\times15^2$$

giving your,

$$272^4 +198593\times 15^4 = 353^4$$

since $k =\pm2$ just factors out. Incidentally, since $198593\times 15^4 = 30^4+120^4+315^4$, then,


More generally, any primitive $(4.1.4)$ can be placed in the form,

$$a^4+(x^4+y^4+z^4)(5m)^4 = b^4$$

II. You gave the identity, $$x^4 + r y^4 = z^4$$ where, $$\begin{aligned} x &= a^4 - 6a^2 b^2 + b^4\\ y &= a^2 + b^2\\ z &= 4a b(a^2 - b^2)\end{aligned}$$ $$r = -(a^8 - 28a^6 b^2 + 70a^4b^4 - 28a^2 b^6 + b^8)$$

The polynomials looked familiar, until I remembered they were in the expansion of the complex number,

$$(a+b\,i)^8 = -r + 2x z\,i$$

hence these particular ($r, x,y,z$) also obey,

$$(a^2+b^2)^8 = r^2 + (2x z)^2 = y^8$$

  • $\begingroup$ Thank you so very much! I find it fascinating how I’ve serendipitously stumbled upon such deep facts. $\endgroup$ – Old Peter Jan 9 '18 at 20:18

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