Is there always a square between two consecutive cubes? Is there always a square between two consecutive cubes? 
I just thought of this question. It seems really simple and the answer is probably yes.
Edit: I should have given this more than 2 seconds of thought.
 A: We show that if $a^3$ and $(a+1)^3$ are two consecutive positive cubes, then there is a perfect square strictly between $a^3$ and $(a+1)^3$. Let $x=\sqrt{a^3}$. If $x$ is an integer, let $b=x+1$. If $x$ is not an integer, let $b=\lceil x\rceil$ (the ceiling function). We will show that $b^2\lt (a+1)^3$.
We have $b\le a^{3/2}+1$. It follows that 
$$b^2\le a^3+2a^{3/2}+1.$$
Now it is enough to show that $2a^{3/2}\le 3a^2+3a$. This is clear.
A: We have $$(n+1)^2-n^2=2n+1$$ and $$(n+1)^3-n^3=3n^2+3n+1.$$ Let $f(x)=2x^{1\over 2}+1$ and $g(x)=3x^{2\over 3}+3x^{1\over 3}+1$. Then:


*

*When do we have $f(x)>g(x)$? (HINT: look at $h(x)=g(x)-f(x)=x^{1\over 3}(3x^{2\over 6}-2x^{1\over 6}+3)$.)

*So?
A: 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.
