Is there any prime $p$ such that $6x^3 − p^2 − y^2 = 0$ has an integer solution?

Question

I need to find whether there is any prime for which $6x^3 − p^2 − y^2 = 0$ has a integer solution.

For prime $p \neq 3$ ,considering this equation in modulo $3$ ,I find that there is no solution.

But for $3$ whether there is a solution or not I could not get this.

Please help.

lulu · Accepted Answer · 2024-02-28T15:43:32.173

4

For $p=3$ we have:

$$6x^3-9-y^2=0\implies 3\,|\,y$$

Writing $y=3Y$ we then have $$6x^3-9-9Y^2=0\implies 2x^3-3-3Y^2=0\implies 3\,|\,x$$ Writing $x=3X$ we then have $$54X^3-3-3Y^2=0\implies 18X^3-1-Y^2=0$$

And now working $\pmod 3$ shows that there are no solutions.

edited Feb 28 '24 at 15:43

answered Feb 28 '24 at 14:56

lulu

76,951

Typo. Last line should read $18X^3$. But this is of no concern. – suckling pig Feb 28 '24 at 15:09
@metoo What do you mean? Dividing by $3$ reduces the $18$ to a $6$, or am I missing something? – lulu Feb 28 '24 at 15:33
Well, you cube $3X$ right? So on the last line you'll have $54X^3-3-3Y^2=18X^3-1-Y^2.$ But still $18$ is zero $\pmod3.$ Nice solution. – suckling pig Feb 28 '24 at 15:38
@metoo Oh, of course. My blunder. That's for pointing it out...I'll edit. – lulu Feb 28 '24 at 15:43

Is there any prime $p$ such that $6x^3 − p^2 − y^2 = 0$ has an integer solution?

1 Answers1