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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?

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    $\begingroup$ 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. $\endgroup$ – quasi Mar 25 '18 at 15:05
  • $\begingroup$ To get a sense of the elliptic curve approach, take a look at: mathworld.wolfram.com/EllipticCurve.html $\endgroup$ – quasi Mar 25 '18 at 15:16
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from $3m^2 + 1 = 4 w^3$ we get the Mordell curve $$ (36m)^2 + 432 = (12w)^3 $$

These are compiled in lists, currently not loading online

However:

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:~$
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  • $\begingroup$ Did not understand the last part. What you meant by "currently not loading"? $\endgroup$ – シャシュワト Mar 25 '18 at 16:08
  • $\begingroup$ Can you explain the output of the computation you did? I really do not understand this. $\endgroup$ – シャシュワト Mar 25 '18 at 20:13
  • $\begingroup$ @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. $\endgroup$ – Will Jagy Mar 25 '18 at 20:17
  • $\begingroup$ 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/… $\endgroup$ – シャシュワト Mar 25 '18 at 20:23

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