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Ahmed Al-Rashid
May 18, 2026

For which primes p does x³ + y³ = p z³ admit non-trivial integer solutions?

Consider the Diophantine equation

x3+y3=pz3,x^3 + y^3 = p z^3,x3+y3=pz3,

where ppp is a prime and we seek non-trivial integer solutions (x,y,z)∈Z3∖{(0,0,0)}(x,y,z) \in \mathbb{Z}^3 \setminus \{(0,0,0)\}(x,y,z)∈Z3∖{(0,0,0)}.

By a classical descent argument using the fact that Q(−3)\mathbb{Q}(\sqrt{-3})Q(−3​) has class number 111, one can show that the equation x3+y3=z3x^3 + y^3 = z^3x3+y3=z3 has only trivial solutions (Fermat's Last Theorem for n=3n=3n=3). For which primes ppp does the generalized equation admit non-trivial solutions?

I suspect that p≡1(mod3)p \equiv 1 \pmod{3}p≡1(mod3) and p=3p = 3p=3 are candidates. Can anyone provide a characterization or refer to known results on cubic forms with prime coefficients?

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1 Answer

3
Ahmed Al-Rashid
May 18, 2026
The proposed characterization by $p \bmod 3$ is not correct as stated. A direct counterexample is $p=2$: $$1^3+1^3=2\cdot 1^3.$$ So primes congruent to $2 \pmod 3$ cannot be excluded. The examples previously given for $p=3$ and $p=13$ were arithmetic errors: $1^3+1^3=2$, not $3$, and $3^3+(-2)^3=19$, not $13$. Factoring in the Eisenstein integers $\mathbb Z[\omega]$ is relevant, but splitting or inertness of $p$ alone does not prove the claimed if-and-only-if statement. A complete characterization of the primes represented by this cubic equation requires additional local and global conditions. Until such a characterization is supplied with a reliable reference, the safe conclusion is that the original congruence claim is false, with $p=2$ as the smallest counterexample.
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