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Let $n\ge 1$. Determine if the equation $x^n+y^n+z^n+w^n=u^{n+1}$ has infinitely many solutions in distinct integers. If so, determine if there are two solutions $(x_i,y_i,z_i,w_i,u_i)$ for $i=1,2$ so that there is no integer a so that one of the solutions equals $(a^{n+1}x,a^{n+1}y,a^{n+1}z,a^{n+1}w,a^n u),$ where $(x,y,z,w,u)$ is the other solution. For instance, the tuples $(1,2,4,5,6)$ and $(1,2,4,7,8)$ satisfy the latter property, and clearly are not solutions for every n.

First of all, note that given a solution $(x,y,z,w,u)$, by setting $(x',y',z',w',u') = (a^{n+1}x,a^{n+1}y,a^{n+1}z,a^{n+1}w,a^n u)$ for any integer a, we obtain another solution (and $(x,y,z,w,u)$ are all distinct iff the elements of the new tuple are distinct). For the given problem, a trivial solution is $(x,y,z,w,u) = (4,4,4,4,4).$ So one gets infinitely many solutions by the initial remark. One approach for solving diophantine equations is to fix all variables except one and regard the expression as a polynomial in the remaining variable. This is particularly useful if the latter polynomial is a quadratic, as the other root can be determined using Vieta's formulas. However for the given problem, this trick doesn't seem to be that useful. For finding infinitely many solutions to diophantine equations, it can be useful to specialize certain variables to make it easier to find solutions. For instance, for the given question, it might be easier to assume $y=x+1, w=z+1$ or to eliminate two variable from consideration somehow. Also, Pell's equations, which are of the form $x^2 - dy^2 = 1$ for some nonsquare integer d, can sometimes be useful for proving that there are infinitely many solutions. If the unique smallest unit $u=x+y\sqrt{d}$ exceeding 1 of $\mathbb{Z}[\sqrt{d}]$ satisfies $x^2-dy^2=1$, then all solutions of Pell's equation are generated by powers of the minimal solution (more precisely all solutions $(x_n,y_n)$ satisfy $x_n+y_n\sqrt{d} = (x+y\sqrt{d})^n$ for some integer n).

Another technique is to use induction. In the case where $n=1$, we need to find four distinct integers that sum to a square of a number distinct from all four integers. We have $1+2+5+8 = 4^2$ works. For $n=2,$ we have $(-1)^2 + 1^2 + (-3)^2 + 4^2 = 3^3.$ Negating all terms other than $u$ (the one on the RHS) won't yield a second solution satisfying the second part of the question though.

Note that if $n$ is odd, then we have the solutions $(0,a,-a,b^{n+1},b^n)$ for any integers $a$ and $b$ so that all four terms are distinct. For instance, we can choose $a=1$ and $b\ge 2$. The latter claim shows the second part holds.

But what if $n$ is even?

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    $\begingroup$ Of course for $n=1$ there are as many solutions as you like. Let $u$ be pretty much anything, say, $u=10$; then you're just asking to write $100$ as a sum of four distinct integers, none equal to ten. Well, $100=1+2+3+94=1+2+4+93=\cdots=1+2+48+49=1+3+4+92=\cdots$. And since any integer is a sum of four squares (and any sufficiently large integer is a sum of four distinct squares), there will be lots of solutions in the $n=2$ case, as well. $\endgroup$ Commented Oct 30, 2022 at 2:14

2 Answers 2

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Given,

$$x^n+y^n+z^n+w^n=u^{n+1}$$

I. For $n=4$

$$x^4+y^4+z^4+w^4=u^5$$

Seiji Tomita has given a parametric solution. His web page link is given below. Click on computational, go to section on fourth powers & click on article 186. http://www.maroon.dti.ne.jp/fermat/eindex.html

$x=-37m^{10}-110nm^9+1170n^2m^8+4440n^3m^7+2310n^4m^6-6552n^5m^5-7770n^6m^4 -1320n^7m^3+1170n^8m^2+370n^9m+11n^{10}$

$y=11m^{10}-260nm^9-1665n^2m^8-1320n^3m^7+5460n^4m^6+9324n^5m^5+2310n^6m^4 -3120n^7m^3-1665n^8m^2-110n^9m+26n^{10}$

$z=-26m^{10}-370nm^9-495n^2m^8+3120n^3m^7+7770n^4m^6+2772n^5m^5-5460n^6m^4 -4440n^7m^3-495n^8m^2+260n^9m+37n^{10}$

$w=19(m^2+nm+n^2)^5$

$u=19(m^2+nm+n^2)^4$

For example, (m,n)=(1,0) we get $(x,y,z,w,u)=(37,11,26,19,19).$


II. For $n=3$

$$x^3+y^3+z^3+w^3=u^4$$

One small numerical solution is $(x,y,z,w,u)=(2,6,7,9,6)$

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$x^n+y^n+z^n+w^n=u^{n+1}$

For, n=2 we have:

$x^2+y^2+z^2+w^2=u^3$

We take:

$(x,y,z,w)=[(m^2-n^2),2(m^2-n^2),(2mn),(4mn)]$

We get:

$u^3=5(m^2+n^2)^2$

To make (RHS) a cube we take, $(m^2+n^2)=(5)^p$

Where, $p=1,4,7,10,13,$----

For p=1, $(m,n)=(2,1)$, $(x,y,z,w,u)=(3,6,4,8,5)$

For $p=4$, $(m,n)=(24,7)$, $(x,y,z,w,u)=(527,1054,336,672,625)$

For $p=7$, $(m,n)=(278,29)$ , $(x,y,z,w,u)=(76443,152886,16124,32248,78125)$

For $p=10$, $(m,n)=(3116,237)$,

$(x,y,z,w,u)=(9653287,19306574,1476984,2953968,9765625)$

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    $\begingroup$ Well, we already know every positive integer can be written as a sum of four squares. $\endgroup$ Commented Nov 3, 2022 at 11:49

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