How do I prove that $$a^3 - b(b+1) = (a-2)(c^2 + 1) + 2$$ has infinitely many solutions if a, b and c are natural numbers?
I have opening the brackets and moving all the terms to one side which gets rid of the constant 2. I have also tried substituting the value of a in b, b in c, bc in a and so on... but it all results in a big mess which I can't simplify.
Ad-hoc solution: note that when $a=2$, $c$ doesn't matter. Solving $$ a^3-b(b+1)=2 $$ for $b$ (given that $a=2$) gives us $b=2$ as well. Then, any triple $(2,2,c)$ $(c\in\mathbb{N})$ is a solution. There are infinitely many such triples so the claim follows.
Better solution: solving for $c$ gives you $$ c=\sqrt{\frac{a-a^3+b+b^2}{2-a}} $$ so to make $c$ a natural number, it is enough that $a$ and $b$ satisfy $$ b+b^2=2a^2-a^3\tag{i} $$ because this will make the numerator under the square root $(2-a)a^2$. Solve (i) for $b$, which yields $$ b=\frac{1}{2}(-1+\sqrt{1-4a+8a^2})=\frac{1}{2}\left(-1+\sqrt{(2a)^2+(2a-1)^2}\right). $$ So we are done if we can demonstrate that there are infinitely many natural numbers $a$ such that $(2a)^2+(2a-1)^2$ is a perfect square. But that is already done here by Hagen von Eitzen. For completeness, I partially reproduce that answer here: define $$ a_0=0, \qquad a_1=3, \qquad a_{n}=6a_{n-1}-a_{n-2}+2\text{ for }n\ge 2. $$ Then, $a_n+(a_n+1)^2=c_n^2$ where $$ c_0=1, \qquad c_1=5, \qquad c_{n}=6c_{n-1}-c_{n-2}\text{ for }n\ge 2. $$ The numbers $\{a_n\}$ alternate in parity so in our context, you want $2a=a_{2k+1}+1$ for $k\geq 1$ (when $k=0$, $a=2$ so refer to the ad-hoc solution above).
Try $(a,b,c) = (-w,x-w,w)$. You get $w^2 + 2w(x+1) = x(x+1)$. How did I do it? I tried $b=a+x$ and $c=a+y$ and tinkered from there to get a simpler equation to deal with that would cancel out higher order terms.
