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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1$\begingroup$ 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. $\endgroup$– quasiMar 25, 2018 at 15:05
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$\begingroup$ To get a sense of the elliptic curve approach, take a look at: mathworld.wolfram.com/EllipticCurve.html $\endgroup$– quasiMar 25, 2018 at 15:16
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$\begingroup$ @quasi he said that m is an odd number greater then 1 so $(m,x)=(1,1)$ can't be a solution $\endgroup$– Chess playerNov 25 at 23:39
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$\begingroup$ @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. $\endgroup$– quasiNov 26 at 0:04
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$\begingroup$ 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 $\endgroup$– Chess playerNov 26 at 6:35
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1 Answer
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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:~$
answered Mar 25, 2018 at 15:45
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$\begingroup$ Did not understand the last part. What you meant by "currently not loading"? $\endgroup$ Mar 25, 2018 at 16:08
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$\begingroup$ Can you explain the output of the computation you did? I really do not understand this. $\endgroup$ Mar 25, 2018 at 20:13
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$\begingroup$ @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. $\endgroup$ Mar 25, 2018 at 20:17
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$\begingroup$ 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/… $\endgroup$ Mar 25, 2018 at 20:23