Does this Diophantine cubic have solutions?

Question

I'm interested in whether $$ 2x^3-1=z^3 $$ has any positive integer solutions. It has solutions mod $m$ for $m\le10^7$ so it seems unlikely that modular means would suffice to prove that there are not solutions. On the other hand I can't find any solutions for $x<10^9$ either.

I'm sure this is a routine problem but I don't know the techniques. Help?

Application: I was reading about near-Fermat triples $x^3+y^3=z^3+1$ where the restriction $x<y<z$ was given rather than $x\le y<z$ and I wondered if the excluded case was possible.

Answer 1

Euler gave an elementary proof that there are no non-trivial solutions. (Actually, he solved the more general equation $y^3+z^3=2x^3$.)

Having asked this before (see Rational solutions of $x^3+y^3=2$) I was pointed to Chapter II, pg 78 here: http://matwbn-old.icm.edu.pl/kstresc.php?wyd=10&tom=42&jez=en.

If you are interested to (partially) solve the equation yourself, the substitutions made in the beginning might allow you to proceed the proof.

Answer 2

Some points:

Greatest common divisor: If $g=gcd(x,y)$, from $1=2x^3-z^3$ we know that $g^3|1$, or $g=1$ and $x$ and $y$ are relative prime.

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Parity: Since $2x^3$ is even and 1 is odd, $z^3$ is odd and $z$ is odd.

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Factors of z: If $p|z$, so $p^3|z^3=2x^3-1$ hence $2x^3\equiv 1 \pmod {p^3}$. Computer search shows $y$ is not dividable by any of $\{2, 3, 7, 13, 19, 37, 61, 67, 73, 79, 97\}$.

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Solving mod primes using computer: From this I found some results that may help in solving or brute-forcing the problem: $x\equiv 0,1,2 ,4 \pmod7$, $x\equiv 0,1,3,9 \pmod{13}$, $x\equiv 0,1,7,11\pmod{19}$,