What is your proof that w^2 + x^3 + y^4 = z^5 has infinite solutions in natural numbers w, x, y, z?

Question

Showing function(s) which actually construct an infinite sequence of such solutions would be optimal.

math_kp:
I do not see the logic in your proof.
Besides which, it is not possible that:
w^2 = z^5 = y^4 = x^3 and my equation holds. Can you provide just one set of natural numbers with your line of reasoning?

Answer 1

O.K., Look for solutions of the form: w^2=x^3=y^4. Making each a power of 3 will cause both sides of the equation to be a power of 3:

3 * 3^n = 3^(n+1) = z^5

where: w = 3^(n/2), x = 3^(n/3), y = 3^(n/4), and z = 3^((n+1)/5)

This is a solution exactly when n is a multiple of 12 and n+1 is a multiple of 5, that is when there's an integer k such that:

n = 12k and 12k + 1 = 0 (mod 5)
-->
k = 2 (mod 5) = 5j + 2
-->
n = 60j + 24
-->
w = 3^(30j+12)
x = 3^(20j+8)
y = 3^(15j+6)
z = 3^(12j+5)

*****

A similar way of generating families of solutions would be to start from any one solution w_0,x_0,y_0,z_0 and define:

w_k = a^(30k)*w_0
x_k = a^(20k)*x_0
y_k = a^(15k)*y_0
z_k = a^(12k)*z_0

so that each of the four terms will be multiplied by a^(60) to go from one solution to the next. Maybe this is what the previous responder thinking. For example 2,3,1,2 is one solution so a family of solutions could be:

w_k = 2*a^(30k)
x_k = 3*a^(20k)
y_k = 1*a^(15k)
z_k = 2*a^(12k)

where a is any integer.

Answer 2

We need to show that
w^2 = z^5 - y^4 - x^3 is a perfect square.

Though I cannot generalize this but for a simple case presume

w^2 = z^5 = y^4 = x^3
this shall meet the criteria atleast set of solutions

take gcd of 5 4 3 that is 60

if w = a^30 then a^60 = w^2
if z = a^5 then a^60 = z^5
if y = a^15 then a^60 = y^4
if z = a^12 then a^60 = z^5
by variying a we get infinite number of solutions