The Diophantine equation $x_1^6+x_2^6+x_3^6=z^2$ where exactly one $(x_i)\equiv 0{\pmod 7}$.

Apart from the trivial $a^6+0^6+0^6=(a^3)^2$, primitive solutions seem difficult to find.

That’s primitive as in not of the form $(kx_1,kx_2,kx_3,k^3z)$ where $(x_1,x_2,x_3,z)$ is a smaller solution.

As far as I can see, either one or two of $(x_1,x_2,x_3)\equiv 0{\pmod 7}$

IMHO, the two cases are very different.

This question concerns the case where exactly one $(x_i)\equiv 0{\pmod 7}$.

My primitive solutions

$$(42,100,81,1134865)$$ $$(168,90,85,4836493)$$ $$(350,324,207,55441585)$$ $$(140,390,213,60163597)$$ $$(378,369,278,76831633)$$ $$(924,1230,715,2053967149)$$

I’ve tried fitting my numerical solutions into known parametric solutions to $a^2+b^2+c^2=d^2$, but without gaining insight. For examples of parametric solutions to $a^2+b^2+c^2=d^2$, see https://sites.google.com/site/tpiezas/004

I know the sixth power residuals ${\pmod 7}$, ${\pmod 8}$ and ${\pmod 9}$ are $0$ or $1$ and that $z\equiv 3{\pmod 7}$ or $z\equiv 4{\pmod 7}$

My Question

Can anybody please do any of the following:

Fit my solutions into the known parametric solutions to $a^2+b^2+c^2=d^2$?

Find a parametric solutions that gives new primitive solutions?

Point me in the direction of any previous work on $x_1^6+x_2^6+x_3^6=z^2$ ?

Find any more primitive solutions?

Update 28th April 2018

The answer from @Sam provides these results, plus others too large for me to easily check. However, it certainly does not provide all the small solutions, so I’m sure there are more to find.

$$(2184,2518,2043,20883327517)$$ $$(3087,4482,3404,102604114673)$$ $$(5306,10617,4728,1210664898377)$$ $$(29316,13469,5802,25313949479269)$$ $$(79758,87036,36221,833297083257349)$$ $$(502026,462741,29408,160707356499029581)$$

• see Robert Spira, The American Mathematical Monthly, Vol. 69, no. 5, May 1962, pages 360-365: The Diophantine Equation $x^2 + y^2 + z^2 = m^2$ – Will Jagy Apr 27 '18 at 19:49
• also see en.wikipedia.org/wiki/… – Will Jagy Apr 27 '18 at 19:55

Equation: $x^6+y^6+z^6=w^2$ ----(1)

Regarding the subsequent request by @OldPeter yesterday, please see below:

Refer to Andrew Bremner & M. Ulas 2011 paper in the International journal of number theory, pages 2018-2090, vol. 8, No 07, having title $(x^a±y^b±z^c±w^d=0)$

The paper includes additional numerical solutions to equation (1) above:

$(x, y, z, w )$

$694, 945, 1308, 2414891825$

$42, 873, 3596, 46505412377$

$792, 3759, 5038, 138465240337$

$1515, 3262, 5160, 141747483853$

$2975, 4950, 7902, 508783710817$

$4410, 5463, 8270, 594854319097$

$5340, 6626, 9765, 987341285501$

$1689, 10528, 14886, 3498954949801$

$588, 8224, 26097, 17782152244433$

$834, 17094, 21373, 10966834991269$

$1182, 14644, 24597, 15209227541197$

• Thank you very much for your answer; I’ve upvoted as these results are very useful. I’m not sure how I can easily see the paper you quote, perhaps you could let me know. I hate to ask, but please will you only refer to me as OldPeter in future. – Old Peter Apr 29 '18 at 13:46

Above equation shown below:

$x^6+y^6+z^6=w^2$

There are two more numerical solutions for $(x,y,z)< 5000$ $(x,y,z,w)=(2043,2184,2518,20883327517)$

$(x,y,z,w)=(3087,3404,4482,102604114673)$

Above solutions including numerical solution shown by "OP" $(x,y,z,w)=(140,213,390,60163597)$

have been arrived at through elliptical curve method by Seiji Tomita on his web site shown below:

  http://www.maroon.dti.ne.jp/fermat/eindex.html


Click on the above link & select 'Computional number theory" &

check out his articles # 166 and #167

• Thank you very much for your answer, which I’ve upvoted. I’m also interested in larger solutions; are these marked “P1:”, “P2:”, “P2-P1:”, “P1+P2:” in the link? – Old Peter Apr 28 '18 at 14:32

Here is a near-solution. If you have one primitive solution to,

$$(2a)^\color{red}2+b^6+c^6 = (2d)^2$$ you can find an infinite more using the identity,

$$(a x^6 - d x^6 + a y^6 + d y^6)^\color{red}2 + (b x y)^6 + (c x y)^6= (a x^6 - d x^6 - a y^6 - d y^6)^2$$

for arbitrary $x,y$.

Example: Given,

$$(81^3)^2+42^6+100^6 = 1134865^2$$

Hence $a=81^3/2$ and $d = 1134865/2$. Using $x= 2,\,y=1$, then,

$$18476415^2 + 84^6 + 200^6 = 20142721^2$$

and infinitely many more where at least one of the $x,y$ is odd.

P.S. It tantalizes there may be a similar identity if $\color{red}2$ is raised to $6$.

• Thanks for your answer. Sorry, apart from $84$ and $200$, I can’t reproduce the figures from your example. I expect it’s me, but could there be any typos in the identity? – Old Peter Apr 27 '18 at 15:49
• @OldPeter: I rechecked it and I get the same numbers. I believe you may have used a different $a$ and $d$, so I've added it explicitly. – Tito Piezas III Apr 27 '18 at 16:05
• Thanks for looking, but we agree on $a$ and $d$. IMHO, the RHS < squared term on LHS. – Old Peter Apr 27 '18 at 17:49
• Tito, your discussion of Pythagorean quadruples at the site the OP linked is misleading: the full parametrization, surely known to Euler but linked with V. A. Lebesgue for a later article, is en.wikipedia.org/wiki/… – Will Jagy Apr 27 '18 at 19:42
• see Robert Spira, The American Mathematical Monthly, Vol. 69, no. 5, May 1962, pages 360-365: The Diophantine Equation $x^2 + y^2 + z^2 = m^2$ – Will Jagy Apr 27 '18 at 19:48

here is the correct parametrization of primitive Pythagorean quadruples 