Proof that $\exists \space a, b, c, \in \mathbb{N} (a,b,c: (a\neq b\neq c)\land (a^2 + b^2 = c^2))$
Today my teacher gave me homework to proof: $$\not \exists\space a, b, c \in \mathbb N (a,b,c: (a \neq b \neq c)\land(a^3 + b^3 = c^3))$$
As far of now, I'm just fimilar with elementry algebra. So, my way was to proof that $a^2 + b^2 + c^2$ exist (see question.) And I'll use the same method to produce a very informal proof that $a^3 + b^3 = c^3$ doesn't exist!
My try:
$$ \forall a, b, c \in \mathbb N , \\ a + b \in \mathbb N \\ \Rightarrow (a+b)^2 \in \mathbb N \\ \Rightarrow a^2 + b^2 = (a+b)^2 - 2ab \\ c^2 = (a+b)^2 - 2ab \space\text{ (assumption)}\\ c^2 - \{ (a+b)^2 - ab\} = 0 \\ c = \frac{\pm \sqrt{4((a+b)^2 - 2ab)}}{2} \\ \Rightarrow c = {\pm\sqrt{(a+b)^2 - 2ab}}$$ Now, if I somehow get a way to get natural expression for $c$ instead of irrational number or a complex number, then I can go for the $3$
Thanks :)
For example, $$(a,b,c)=(3,4,5)$$ because $$3^2+4^2=5^2,$$ but there are infinitely many these triples:
$$(a,b,c)=(2mn,m^2-n^2,m^2+n^2)$$ because $$(2mn)^2+(m^2-n^2)^2=(m^2+n^2)^2.$$ Check it.