When is the difference of two consecutive positive cubes a perfect square?

Question

Are there only finitely many solutions in positive integers $m,n$ to the equation $$(m+1)^3-m^3=n^2\; ? $$

Answer 1

There are infinitely many of them. Let

$$f(m,n) \stackrel{def}{=} (m+1)^3 - m^3 - n^2$$ It is clear the equation $f(m,n) = 0$ has a trivial solution $(m,n) = (0,1)$.
By brute force, one can verify

$$f(7m+4n+3,12m+7n+6) = f(m,n)$$

This means if $(m,n)$ is a solution for $f(m,n) = 0$, so does $$(m',n') = (7m+4n+3,12m+7n+6).$$ Start from the trivial solution $(0,1)$, we can use this to generate infinitely many positive integer solutions for the equation $f(m,n) = 0$.

Answer 2

You have the quadratic equation $3m^2+3m+1=n^2$, which has positive discriminant. In general, every quadratic equation in two variables with a positive discriminant either has $0$ or $\infty$ solutions. Since there is at least one (namely $(m,n)=(0,1)$), there are infinitely many.

You may want to read about Pell's equation $x^2-dy^2=1$ or the more general (sometimes called Pell-type equation) $x^2-dy^2=a$ to learn why this kind of equations either has $0$ or $\infty$ solutions. In fact, by studying the Pell-type equation associated to a particular such quadratic it is possible to give explicit recursive formulas to construct solutions (see achille hui's answer here). Moreover, all solutions are given by a finite number (that can be arbitrarily large, though it's always possible to give an upper bound) of such recursive families.

Answer 3

We write the equation:

$$(x+1)^3-x^3=y^2$$

$$3x^2+3x+1=y^2$$

Let's use the formula. https://mathoverflow.net/questions/31118/integer-polynomials-taking-square-values/195614#195614

For these equations we use the standard approach. For a private quadratic form: $$y^2=ax^2+bx+1$$

Using solutions of Pell's equation: $$p^2-as^2=1$$

Solutions can be expressed through them is quite simple.

$$y=p^2+bps+as^2$$

$$x=2ps+bs^2$$

$p,s$ - these numbers can have any sign.

Finding solutions of equations Pell - standard procedure.

So for our case. Use the Pell equation. $p^2-3s^2=1$

And then the decision will have a look.

$$x=2ps+3s^2$$

$$y=p^2+3ps+3s^2$$

Knowing the first solution. $( p_1 ; s_1 ) - (2 ; 1 )$

You find the following.

$$p_2=2p_1+3s_1$$

$$s_2=p_1+2s_1$$

Be aware that the number $(p;s)$ can have any sign.