Proving Non-Equality of Cubes of Natural Numbers

[mathmaniac1]

It's quite trivial to prove that.

Absolutely!My understanding

Let $$b^2$$ be an odd square number.There exists a pythagorean triple including $$b^2$$ because the odd number $$b^2$$ can form the last number of the string of consecutive odd numbers of another square number $$a^2$$.So there are PPTs as long as there are odd numbers and their squares,so they are infinite...

 

[mathworker]

i have a little different approach to prove there are infinite triples ,(though its unnecessary)
let x,z be two variables ,
let x^2+z^2=(x+y)^2
then z^2=y^2+2xy =y(y+2x)
and if we consider y to be perfect square we can adjust x such that y+2x is perfect square
since there are infinite perfect squares there and infinite x's and thus we can find infinite z's thus there are infinite solutions to x^2+y^2=z^2

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i tried applying same method to n=3 but i found myself cycling around ...