Solving Diophantine equation with parameters Today i want to learn/discuss about of a technique for proving that the given Diophantine equation has infinitely main roots.
I just want to solve these types of problems :
1) Show that $x^2=y^3+z^5$ has infinitely many solutions for positive integers $x,y,z$.
2) Show that $x^n+y^n=z^{n-1}$ has infinitely many solutions for postive integers $x,y,z$.
To solve these types of problems a parameter is usually used which varies over integers giving infinitely many solutions.. For example, $x=k^{10}(1+k)^8 ,y=k^7(1+k)^5,z=k^4(k+1)^3$ are the solutions for problem number (1). Then as we vary $k$ over postive integers we will be getting infinitely many values.
My problem is that I am not getting motivation to how to select that values of $x,y,z$ in terms of $k$ or any parameter. After seeing the solution I feel "Ohk! It can be done using this" but i can't predict the solution.
Hence I am asking is there any procedure are to be followed to solve these types of problems , any motivation inside the question?? Or it can be solved only by putting random value??
Please help me.
Thanks in advance,
 A: One of my friend on Aops @Delta0001 has discovered this method for solving these types of methods of one type. So I want to share that trick to the math community.
Prove that there are infinite tuples of natural numbers $(a_1 , a_2 , \dots , a_n)$ which satisfy
$$a_1 ^{p_1} + a_2 ^{p_2} + \dots + a_n ^ {p_n} = d^p$$
where $p , p_1, p_2 , \dots , p_n , d$ are natural numbers and  $gcd ( ~ lcm( p_1 , p_2 , \dots , p_n ) ~ , ~ p)  = 1$
Solution
Let $M$ denote the LCM of $p_1 , p_2 , \dots , p_n$.
Next, we find $k$ such that $M \cdot k + 1$ is a multiple of $p$.
Now, take any natural numbers $b_1 , b_2 , \dots , b_n$ and set,
\begin{align*}
a_1 &= b_1 (b_1 ^ {p_1} + b_2 ^{p_2} + \dots + b_n ^{p_n}) ^{\frac{M k}{p_1}} \\
a_2 &= b_2 (b_1 ^ {p_1} + b_2 ^{p_2} + \dots + b_n ^{p_n}) ^{\frac{M k}{p_2}} \\
& \vdots \\
a_n &= b_n (b_1 ^ {p_1} + b_2 ^{p_2} + \dots + b_n ^{p_n}) ^{\frac{M k}{p_n}}
\end{align*}
Also, let $d  = (b_1 ^ {p_1} + b_2 ^{p_2} + \dots + b_n ^{p_n}) ^{\frac{M k + 1}{p}}$
It is clearly seen that this tuple $a_1 , a_2 , \dots , a_n , d$ satisfy the given equation.
And as there are infinitely many choices for $b_1 , b_2 , \dots m b_n$, we get infinitely many solutions. $\blacksquare$
A: "OP" inquired about,  $x^n+y^n=z^{n-1}$
For n=3, above equation becomes: 
$x^3+y^3=z^2  ----(1)$
Equation $(1)$ has parametric solution given below:
$x=2(m^2+1)(m^4+3)$
$y=2(m^2-1)(m^4+3)$
$z=4m(m^4+3)^2$
For, $m=3$, we get:
$1680^3+1344^3=(84672)^2$
