Prime exponent Diophantine equations with infinitely many solutions I am to prove that the following equation has infinitely many solutions in the set of natural numbers:
$${x_1}^{p_1} + {x_2}^{p_2} + \cdots + {x_{n-1}}^{p_{n-1}} = {x_n}^{p_n},$$ where $p_1, \dots ,p_n$ are given different prime numbers. 
Should one resort to induction, checking the first step is not so difficult, as the case for $n=1$ is obvious, and as to $n=2$ for the equation ${x_1}^{p_1} = {x_2}^{p_2}$ one could set $x_1= p^{p_2} \, , x_2=p^{p_1}$ for an arbitrary prime $p$. 
But in passing from the induction hypothesis (the case $n$) to the case $n+1$, I could not get through. Maybe, of course, the problem is solved without induction. Thanks for any suggestion or solution! 
 A: The question can be simplified down to: does there exists naturals $a_i$ for $1\leq i \leq n$ such that
$$a_1p_1=a_2p_2=\dots=a_{n-1}p_{n-1}=a_np_n-1$$
This is in fact true. Let $L=p_1p_2\cdots p_{n-1}$. Clearly, $p_i|L$ for $1\leq i\leq n-1$ and that $\gcd\left(p_n,L\right)=1$. Thus, there exists $x,y\in\mathbb{N}$ such that
$$xp_n-yL=1\Rightarrow yL=xp_n-1$$
Now, for $1\leq i\leq n-1$, define
$$a_i=y\frac{L}{p_i}$$
and 
$$a_n=x$$
Then
$$a_ip_i=yL\text{ for }1\leq i \leq n-1$$
and
$$a_np_n=xp_n=yL+1$$
Having proved there exists naturals such that
$$a_1p_1=a_2p_2=\dots=a_{n-1}p_{n-1}=a_np_n-1$$
is true, we move on to the end of the proof. As @John_Omiela noted in his comment, if you have one solution you have infinite solutions. One solution is
$$x_i=(n-1)^{a_i}$$
Then
$$x_1^{p_1}+x_2^{p_2}+\cdots +x_{n-1}^{p_{n-1}}=(n-1)^{a_1p_1}+(n-1)^{a_2p_2}+\cdots+(n-1)^{a_{n-1}p_{n-1}}$$
$$=(n-1)(n-1)^{yL}=(n-1)^{yL+1}$$
while
$$x_n^{p_n}=(n-1)^{a_np_n}=(n-1)^{yL+1}$$
Thus, equality has been established and we are done.
A: Yes, you can always have an infinite number solutions solutions. First, let
$$Q = \prod_{i=1}^{n-1}p_i \tag{1}\label{eq1A}$$
Next, since $Q$ would be relatively prime to $p_n$, then you have
$$kQ \equiv -1 \pmod{p_n} \implies k \equiv -Q^{-1} \pmod{p_n} \tag{2}\label{eq2A}$$
Choose one of these $k$ and then have
$$x_i = \left(n-1\right)^{\frac{kQ}{p_i}}, \; 1 \le i \le n - 1 \tag{3}\label{eq3A}$$
The equation you're asking to solve is
$$\sum_{i=1}^{n-1}x_i^{p_i} = x_n^{p_n} \tag{4}\label{eq4A}$$
The LHS side of \eqref{eq4A} is then
$$\begin{equation}\begin{aligned}
\sum_{i=1}^{n-1}x_i^{p_i} & = \sum_{i=1}^{n-1}\left(\left(n-1\right)^{\frac{kQ}{p_i}}\right)^{p_i} \\
& = \sum_{i=1}^{n-1}\left(n-1\right)^{kQ} \\
& = \left(n-1\right)^{kQ+1}
\end{aligned}\end{equation}\tag{5}\label{eq5A}$$
However, by \eqref{eq2A}, $kQ + 1$ is a multiple of $p_n$, say
$$kQ + 1 = mp_n \tag{6}\label{eq6A}$$
Thus, you can set
$$x_n = (n-1)^{m} \tag{7}\label{eq7A}$$
This shows a set of values which solve \eqref{eq4A}.
As for getting an infinite number of solutions, have
$$R = \prod_{i=1}^{n}p_i \tag{8}\label{eq8A}$$
Next, choose any
$$m \gt 1, \; m \in \mathbb{N} \tag{9}\label{eq9A}$$
Then consider
$$x_i^{'} = \left(m^{\frac{R}{p_i}}\right)x_i, \; 1 \le i \le n \tag{10}\label{eq10A}$$
Each term on the left & right of \eqref{eq4A} would be multiplied by $m^{R}$, so it would still be satisfied.
