Does the Diophantine equation $z_1^5 +z_2^5+z_3^5+z_4^5+z_5^5=\beta^5$ have a solution for every integer $\beta$?

Question

(Note: The exponent $k=3$ has been answered in the affirmative in this post.)

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I. Data

For simplicity, assume all terms $\in \mathbb{Z},$ so we can transform the equation to the more symmetric,

$$x_1^5+x_2^5+x_3^5 = y_1^5+y_2^5+y_3^5$$

where $(x_1, x_2, x_3) = (y_1, y_2, y_3)$ are considered trivial solutions. In 2009, by an exhaustive search, Duncan Moore found roughly 5400 primitive solutions within a search radius of about $17700.$ For example,

\begin{align} 1^5 & + 89^5 + 118^5 = 123^5 + 47^5 + 38^5\\ 2^5 & + 97^5 + 258^5 = 257^5 + 125^5 + 35^5\\ 3^5 & + 54^5 + 62^5 = 67^5 + 28^5 + 24^5\\ 4^5 & + 32^5 + 498^5 = 463^5 + 369^5 + 302^5\\ 5^5 & + 145^5 + 224^5 = 214^5 + 157^5 + 153^5 \\ 6^5 & + 265^5+ 614^5 = 543^5+ 527^5+ 235^5\\ 7^5 & + 201^5+ 303^5 = 307^5+ 173^5+ 31^5\\ 8^5 & + 62^5+ 68^5 \,=\, 74^5+ 43^5+ 21^5\\ 9^5 & + 206^5+ 430^5 = 418^5+ 297^5+ 20^5\\ 10^5 & + 100^5+ 972^5 = 951^5+ 617^5+ 204^5\\ \vdots\\[4pt] 200^5 & + 334^5 + 676^5 = 679^5 + 256 ^5 + 185^5 \end{align}

up to $x_1 = 200$ which is quite a long stretch. But there were three missing: namely $x_1 = (22,\,88,\,176)$, all of which are multiples of $11$ and a power of $2$.

Update: The list for $0\leq x_1\leq 1000$ is now complete, with the last two, namely ($410, 840$), found by Oleg567. See his answer below.

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II. Updates

1. As Adam Bailey pointed out, there doesn't seem to be an obvious congruence obstruction for $x_1 = 22$ and others. So it may be possible for all $x_1$ just like its cousin $z_1^3+z_2^3 = z_4^3+z_4^3$ (though by FLT, this has no $z_1 = 0$).
2. There is $0^5 + 220^5 + 14132^5 = 14068^5 + 6237^5 + 5027^5$, so $x_1=0$ is now possible for $5$th powers.
3. Oleg567 found $x_1=176$ valid for $k=1,5$ using a larger search radius, $$176^5+20117^5+22952^5=5781^5+12692^5+24772^5$$ $$176+20117+ 22952=5781+12692+24772$$
4. Moore found $x_1=22$ also valid for $k=1,5.$ (See his answer below.)
5. Using the form ($5,2,4$), wxffles found $x_1=88$ and $x_1=858$ valid only for $k=5.$ (See addendum to Moore's answer for $x_1<1000$.)
6. James Waldby's database for ($5,1,5$) can be found here for other missing $x_1$.

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III. Question

Can we in fact find a primitive solution $\in \mathbb{Z}$ for any integer $x_1$,

$$x_1^5+x_2^5+x_3^5 = y_1^5+y_2^5+y_3^5$$

hence the absence of $x_1 = 22, 88, 176,$ etc. is simply an artifact of the search radius?

Answer 1

This is a partial answer. There is a solution to $x_1 =22$, namely,

$$22^5+42303^5+49013^5=51233^5+36563^5+3542^5$$ $$22+42303+49013=51233+36563+3542$$

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I. Extended searches

In 2017 I extended my previous exhaustive search (2009) of $x_1^5+x_2^5+x_3^5=y_1^5+y_2^5+y_3^5$ to a radius of $52586$ (i.e. the sum searched to is $52586^5$). There are 15904 solutions and more than half (54.6%) also satisfy $x_1+x_2+x_3=y_1+y_2+y_3$.

In 2017/2018 I ran a separate exhaustive search of the same equation under the constraint $x_1+x_2+x_3=y_1+y_2+y_3\le205848$, giving 16460 solutions, many of which are larger than those in the first set.

The first extended search eliminates 176, as already found by Oleg567. The second search eliminates 22.

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II. Statistics

Given $x_1^k+x_2^k+x_3^k=y_1^k+y_2^k+y_3^k$, statistics for the first extended search are,

$$\begin{array}{|c|c|c|c|c|} \hline \text{Radius} & \text{Solns k = 5} & \text{Solns k = 1,5} & \text{% of k = 1,5} & \text{% diff} \\ \hline 13146 & 3957 & 2378 & 60.1 & --\\ \hline 26293 & 8108 & 4657 & 57.4 & 2.7\\ \hline 39439 & 11982 & 6683 & 55.8 & 1.6\\ \hline 52586 & 15904 & 8677 & 54.6 & 1.2\\ \hline \end{array}$$

To explain, for the first radius of 13146, then $2378/3957 = 0.601$, or 60.1% of solutions (solns) are valid for two exponents $k=1,5$. For the highest radius of 52586, this falls to 54.6%. However, the decline (% diff) seems to be slowing down, so whether this converges to some percentage is unknown.

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III. Addendum (by Piezas)

In Moore's ($5,3,3$) database, there are only 9 missing terms with $x_1<1000$: $$88, 275, 410, 495, 785, 800, 840, 850, 858$$ But by adding the databases of wxffles, Waldby, and an identity by Sastry, we find,

$$\color{blue}{88}^5 + 367^5 = 335^5 + 298^5 + 140^5 + 132^5$$ $$\color{blue}{495}^5 + 1043^5 = \color{blue}{858}^5 + 864^5 + 795^5 + 71^5$$ and, $$\color{blue}{275}^5 + 351^5 + 872^5 + 1298^5 + 1855^5 = 1921^5$$ $$\color{blue}{495}^5 + 218^5 + 276^5 + 385^5 + 409^5 = 553^5$$ $$\color{blue}{785}^5 + 1023^5 + 2782^5 + 5407^5 + 5500^5 = 6287^5$$

with the first two from wxffles's ($5,2,4$) database, the last three from James Waldby's ($5,1,5$) database, while Sastry's 1934 parameterization,

$$(\color{blue}{50 n})^5 + (10 n^3)^5 + (n^5 - 25)^5 + (n^5 + 25)^5 = (n^5 - 75)^5 + (n^5 + 75)^5$$

takes care of $x_1$ that are multiples of $50$. Thus only two $x_1<1000$ remain missing: $410, 840$.

Answer 2

Partial answer :

Use :

$$(x+y+z)^5-x^5-y^5-z^5-5(x+y)(z+y)(z+x)(x^2+z^2+y^2+xy+yz+zx)=0$$

Then solve :

$$x+y+z=u+v+w=0,(x+y)(z+y)(z+x)(x^2+z^2+y^2+xy+yz+zx)=(u+v)(w+v)(w+u)(u^2+w^2+v^2+uv+vw+wu)=k$$

The identity is called Lamé identity found on your site @TitoPiezasIII

Another idea :

Let introduce (all the variable are integers):

$$a(x,y,z,u,v)=(x+y+z+u+v)^5-(x-y+z+u+v)^5-(x+y-z+u+v)^5-(x+y+z-u+v)^5-(x+y+z+u-v)^5+(x-y-z+u+v)^5+(x-y+z-u+v)^5+(x-y+z+u-v)^5+(x+y-z-u+v)^5+(x+y-z+u-v)^5+(x+y+z-u-v)^5-(x-y-z-u+v)^5-(x-y+z-u-v)^5-(x+y-z-u-v)^5-(x-y-z+u-v)^5+(x-y-z-u-v)^5=1920xyzuv$$

Then we can pick three solution as $$a^5=b^5+c^5-u^5-v^5-w^5,A^5=B^5+C^5-U^5-V^5-W^5,0=B'^5+C'^5-U'^5-V'^5-W'^5$$

To get another one (it's a kind of telescoping).

The big identity is called Boutin's identity ($n=5$) found again on the site for TitoPiezasIII

Answer 3

COMMENT.-In relation with the well-known Waring's Problem, the Chinese mathematician Jung-Run Chen has proven, many years ago, that all natural numbers can be represented as the sum of $37$ fifth powers of natural numbers. So if you are right, we have for all natural number $\beta$ with $n_i \in \mathbb N,$ $$\beta^5=\sum_{1\le i\le37}n_i^5$$ and also $\beta$ with $z_i \in \mathbb Z,$ $$\beta^5=\sum_{1\le i\le5}z_i^5$$ The required minimum $37$, necessary to have the result valid for all natural integers, found by Chen, would be denied for all the $5^{th}$powers. There are reasons to be skeptical, despite you work in $\mathbb Z$ where there is greater operability than in $\mathbb N$.

Answer 4

Still partial answer
(completion of the range $0 \le x_1 \le 1000$)

Missing numbers $410, 840$ (and $800$) are covered by such primitive solutions:

$$\color{orchid}{410}^5+581^5+1462^5+3515^5=2622^5+3346^5,$$ $$\color{orchid}{840}^5+800^5+1052^5+1996^5=27^5+2021^5,$$

Note that the example of primitive solution for $800$: $$\color{orchid}{800}^5+40^5+801^5+2399^5=799^5+2401^5,$$ is generated by $2$-parameter Sastry's 1934 parameterization (see https://mathworld.wolfram.com/DiophantineEquation5thPowers.html, formula $(5)$, when $u=1, v=2$).

Some example of primitive solution for $1000$: $$\color{orchid}{1000}^5+5317^5+5393^5+8527^5=690^5+8837^5. $$ A much bigger primitive solution for $1000$ could be derived from the parametrization above (when $u=1, v=10$): $$\color{orchid}{1000}^5 +500000^5+2500001^5+7499999^5 =2499999^5 + 7500001^5.$$