Almost Pythagoras Triplet?

I didn't know this identity, so I approached it in a slightly different way.

$a^2+b^3=c^2\implies c^2-a^2=b^3\implies (c+a)(c-a)=b^3$

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Lemma: Every odd positive integer $x$ can be expressed as $(z-a)(z+a)$ for some integer $a,z$

Proof: Every odd positive integer can be expressed in the form $2i+1$ for some integer $i$ . Take $z=i+1$ and $a=i$ . So, we get,

$(z-a)(z+a)=x=2i+1\\ \implies (i+1-i)(i+1+i)=2i+1\\ \implies 2i+1=2i+1$

This shows that the lemma we have established is indeed an identity for all odd positive integers $x$ and hence the lemma is proved.

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Using this lemma and the fact that $b^3$ is odd for odd $b$ , we can conclude that for every odd $b$ , we always have some integral values of $a,c$ . And since the set of odd positive integers is infinite, we have infinitely many integral triples of $(a,b,c)$ .

By the way, you might consider editing your solution since the infinitely many triples aren't of the form of $(n,(n+1),(n+2))$ .

Take $n=0$ . The triple $(0,1,2)$ doesn't satisfy the equation.

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The triples will be of the form $\left(\dfrac{n(n+1)}{2},n+1,\dfrac{(n+1)(n+2)}{2} \right)$

I did it the same way !! :) .. (But got stuck at the penultimate step :") )