Cubic Diophantine Equation of Two Variables

Question

How can the Diophantine Equation $x^3 - y^3 = n$ be solved where $n$ is a positive integer? Do any restrictions necessarily apply to $n$?

Answer 1

No, it can not be solved for all $n$. If $n$ is prime, then $x-y=1$ and $x^2+xy+y^2=p$, so we get quadratic equation on $y$ with parameter $n$: $$3y^2+3y+1-n=0$$

But it disciriminat is $12n-3$ which is not a perfect square for all prime $n$, say $n=2,3...$.

So for prime $n$ it can be solved nly if $12n-3$ is a perfect square.

Answer 2

Equation, $n=x^3-y^3$ -------- (A)

Solution given by 'Greedoid" is for 'n' equal to a prime.

But'OP' is looking for positive integers including prime numbers. Hence, since $x>y$ we take $(x-y=k)$. and so we get $x^3-y^3=kw=n$. Where $w= (x^2+xy+y^2)$. Thus the discriminant is $(12w-3k^2=m^2)$.

For $(w,k,m)=(7,1,9)$ we get $(x,y,n)=(2,1,7)$

For $(w,k,m)=(7,5,3)$ we get $(x,y,n)=(3,-2,35)$

Hence the condition on equation (A) is:

$n=(k/12)*(m^2+3k^2)$.

Answer 3

There are some modular restrictions on $n$.

For any $m$, $m^3\equiv0,1,-1\pmod7$, so $n\equiv3,4\pmod7$ is impossible.

Also $m^3\equiv0,1,-1\pmod9$, so $n\equiv3,4,5,6\pmod9$ is impossible.