The Diophantine equation $x_1^6+x_2^6+y^6=z^2$ where both $(x_i)\equiv 0{\pmod 7}$.
As a logical follow on to The Diophantine equation $x_1^6+x_2^6+x_3^6=z^2$ where exactly one $(x_i)\equiv 0{\pmod 7}$. I have considered the case with exactly two of the sixth powers $\equiv 0{\pmod 7}$.
Although I’ve attempted two different methods, I’ve not found a solution up to $z=10^{12}$
Method 1
Calculate $(7a)^6+(7b)^6+y^6$ within a range, testing if each result is square.
Method 2
As I’m interested in primitive solutions,
$$x_1=7a$$
$$x_2=7b$$
$$7^6(a^6+b^6)=z^2-y^6=(z-y^3)(z+y^3)$$
Then either $(z-y^3)\equiv 0{\pmod 7^6}$ or $(z+y^3)\equiv 0{\pmod 7^6}$.
Using $f_1=z-y^3$ and $f_2=z+y^3$
when $z=7^6c+y^3$ we have $f_1=7^6c$
when $z=7^6d-y^3$ we have $f_2=7^6d$
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I’ve omitted much of the details of secondary modular tests from the above.
To my surprize, I found both methods ran at about the same speed.
My question
Can anyone find a non-trivial solution or prove any useful constrains (perhaps $z$ must be a cube as a wild guess) please?
Please note that I do not have access to an academic library.
Best Answer
Above equation shown below:
$x_1^6+x_2^6+y^6=z^2 -----(1) $
Re: In the question (OP) is guessing if $(z)$ in equation (1) is a cube. But that will turn equation $(1)$ into a $(k,m,n) = (6-1-3)$ equation. Where degree k=6 and where $(m,n)$ are number of terms on each side of the equation . Since $(m+n=4)$ is less than $(k=6)$ his guess is not doable. Incidentally another numerical solution for equation (1) is $(x,y,z)=(36221,79758,87036)$ for $((x+y+z)>50000)$. This solution is given in "Bremner & Ulas" year $2011$ paper in the International Journal of Number theory, volume 8 # 7. Seiji Tomita mentions the above paper on his webpage, so he might have a copy. His Email info his provided on his webpage.