# Proving that this expression can never be a perfect cube

I want to prove that $\dfrac{3m^2+1}{4}$ can never be a perfect cube. Here $m$ is an odd number greater than $1$. Is there a simple way to do that? I saw other answers proving some expression cannot be perfect squares using modulo operators. Could that be used here in this case?

• If for some positive integer $n$, the congruence $$3m^2+1\equiv 4x^3\;(\text{mod}\;n)$$ had no solutions $(x,m)$, that would prove that the equation $$3m^2+1=4x^3$$ has no integer solutions. Unfortunately, the equation does have integer solutions, for example $(m,x)=(1,1)$, which strongly suggests that you can't get a simple resolution via congruences. Instead, equations of this form can be analyzed by (advanced) elliptic curve methods. Mar 25, 2018 at 15:05
• To get a sense of the elliptic curve approach, take a look at: mathworld.wolfram.com/EllipticCurve.html Mar 25, 2018 at 15:16
• @quasi he said that m is an odd number greater then 1 so $(m,x)=(1,1)$ can't be a solution Nov 25 at 23:39
• @Chess player: If the congruence mod $n$ has no solutions for some $n$, then the associated equation with conditions relaxed to allow $(m,x)=(1,1)$ would also have no solutions. Nov 26 at 0:04
• that is what he wanted to get, that it didn't get any solutions. But I still don't get how you got the answer Nov 26 at 6:35

from $3m^2 + 1 = 4 w^3$ we get the Mordell curve $$(36m)^2 + 432 = (12w)^3$$
jagy@phobeusjunior:~$sage ┌────────────────────────────────────────────────────────────────────┐ │ SageMath Version 6.9, Release Date: 2015-10-10 │ │ Type "notebook()" for the browser-based notebook interface. │ │ Type "help()" for help. │ └────────────────────────────────────────────────────────────────────┘ sage: E = EllipticCurve([0,0,0,0,-432]) sage: E.integral_points() [(12 : 36 : 1)] sage: quit Exiting Sage (CPU time 0m0.50s, Wall time 1m3.94s). jagy@phobeusjunior:~$
• @ShashwatSharan it says the only integral point satisfying $s^2 = t^3 - 432$ is $s=36, t=12$ Meanwhile, the proof of Fermat's Last Theorem for $n=3$ is done very well in Ireland and Rosen. Mar 25, 2018 at 20:17
• Which implies only integer solutions are $(1,1)$. I really thank you for your efforts but what I needed was a theoretical proof because we know Fermat's Last Theorem is true. This had to be the case that the only solution is $(1,1)$. Why I referred to Fermat's Last Theorem is because of my following question. math.stackexchange.com/questions/2707813/… Mar 25, 2018 at 20:23