# Are there useful criterions whether a positive integer is the difference ot two positive cubes?

This question

A Diophantine equation involving factorial

made me try to find a useful structure for the set of positive integers that are the difference of two cubes. I have two questions similar to the linked question :

Define $S$ to be the set of positive integers $N$, that are the difference of two positive cubes (in other words, $N=x^3-y^3$ is solveable in positive integers, note that $y=0$ does not count as a solution)

I wrote a routine in PARI/GP to check whether a number is in $S$ :

cubes(n)={gef=0;w=factor(n);u=component(w,1)~;v=component(w,2)~;forvec(z=vector(length(v),m,[0,v[m]]),t=prod(j=1,length(v),u[j]^z[j]);if((gef==0)*(t^3<n),if(issquare((12*n-3*t^3)/t)==1,gef=1)));gef}


The first elements in $S$ are :

? q=0;for(a=1,1000,if(cubes(a)==1,q=q+1;print1(a," ");if(Mod(q,18)==0,print)))
7 19 26 37 56 61 63 91 98 117 124 127 152 169 189 208 215 217
218 271 279 296 316 331 335 342 386 387 397 448 469 485 488 504 511 513
547 602 604 631 657 665 702 721 728 784 817 819 866 875 919 936 973 988
992 999
?

• Does $S$ contain a number $N$ that has only the prime factors $2,3$ and $5$ ? I checked the numbers $2^a\cdot 3^b \cdot 5^c$ in the range $0\le a,b,c\le 40$ and found none.
• Does $S$ contain a number $N$ of the form $lcm(1,2,3,\cdots ,a)$ ? I checked the numbers upto $a=70$ and again I found none. Removing the cube part does not help much here, in the case of the factorials removing the cube part is more helpful.

Concerning the linked question, no factorial $a!$ upto $a=40$ and no cube free part of a factorial $a!$ with $a\le 107$ is in $S$.

This is very much more a comment than an answer, and relates to just your first question.

We have $N=x^3-y^3$, and want to show that $N$ is not a $5$-smooth number, http://oeis.org/A051037 (also known as the Hamming sequence).

$N=x^3-y^3=(x-y)(x^2+xy+y^2)$

Put $M=x^2+xy+y^2$

My limited testing shows the greatest prime factor of $M>5$, regardless of the value of $x-y$, which I hope is a stronger conjecture, and perhaps easier to prove.

Put $z=x-y$, then $N=z(3y^2+3yz+z^2)$ and $M=3y^2+3yz+z^2$

We can force $z$ to be $5$-smooth with $z=2^F3^G5^H$ where $F,G,H$ are non-negative integers. This gives

$$M=3y^2+2^F3^{G+1}5^Hy+4^F9^G25^H$$

In the simplest case, $F=G=H=0$

$$M=3y^2+3y+1,z=1$$

$$M=3y(y+1)+1$$

As either $y$ or $y+1$ is even, and $M$ is $1$ more than a multiple of $3$, so $M=1mod6$.

By trial, $Mmod5=(1,2,4)$, so $N$ is not $5$-smooth for $z=1$.

Clearly, this just a subset, but perhaps it will help.