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I was trying to prove the Fermat's Last Theorem for the special case of $n=3$ by proof by contradiction. The following goes my thought process. This is kind of a geometric way of thinking.

Assume there exists at least one solution for the equation $x^3+y^3 =z^3$ $\ \ \ \ \ \ \mathbf{(1)}$.

Imagine $2$ cubes one of side length $\mathbf{x}$ and other of side length $\mathbf{y}$ (now on I am going to refer cubes by their side length for e.g. cube of side length $\mathbf{x}$ will be referred as $\textbf{cube$_x$}$). Now divide the cube$_x$ into $\mathbf{x}^3$ unit cubes. Now start placing the unit cubes on the surface of the cube$_y$ in a way such that this process leads us to have $\textbf{cube$_z$}$. By doing this process we would have covered three faces of the original cube$_y$. The volume of the new cube$_z$ formed will be: $$z^3 = y^3 + y^2+y^2+y^2+y+y+y+1 $$ $$\Rightarrow z^3=y^3 + 3y^2+3y+1 \ \ \ \ \ \ \mathbf{(2)}$$ The contribution of $x^2$ comes from the cubes added on the surface of cube$_y$ and each contribution of '$x$' comes from the cubes added on the three edges of the cube$_y$ and finally the contribution of '$1$' comes from the the cube added to the vertex of the intersection all the three edges on which unit cubes were added. This is nothing but the binomial expansion of $(1+$y$)^3$. This implies that the cubes were added in such a way that the thickness of the cubes increased by $1$ unit. The argument could easily be generalized such that after the cube addition process the thickness of the cube increases by say, $\mathbf{k}$ units (k is a positive integer). Then the volume of cube$_z$ will be given as$$ z^3=(y+k)^3=y^3+k^3+3k^2y+3ky^2 \ \ \ \ \ \mathbf{(3)} $$

Now, since we have assumed that there exists at least one solution for the equation $\mathbf{(1)}$, which implies that in $\mathbf{(2)}$ the last part i.e. $3x^2+3x+1$ must be a perfect cube. In general,$k^3+3k^2y+3ky^2$ must be a perfect cube for some $y$ whose value must equal $x^3$ as the volume must be conserved while transformation. $$\Rightarrow x^3 = k^3+3k^2y+3ky^2 \ \ \ \ \ \mathbf{(4)}$$ Rearranging this equation as follows by completing the square.$$ x^3 = \dfrac{k}{4}\left(3(2y+k)^2+k^2\right) \ \ \ \ \ \mathbf{(5)}$$ For now assuming $k=1$.

This will not be a loss of generality as if we assume $\mathbf{(1)}$ to be true for some integer inputs, which we are, which means two cubes could be combined to form a new cube of integer side length, to get to $k=2$, first the case $k=1$ must be covered. In other words, to cover a larger cube of integer side with unit cubes such that each dimension of larger cube increases by, say, $2$ units the cube must be covered such that its each dimension first increases by $1$ unit and then reaches $2$ units. $2$ units cannot be increases without the cube first reaching a point of having its dimensions increased by $1$ unit. And because of $x$ and $y$ being arbitrary if we prove of this $x$ and $y$, we prove for all $x$ and $y$.

So, we get after putting $k=1$, $$x^3 = \dfrac{1}{4}\left(3(2y+1)^2+1\right) \ \ \ \ \ \mathbf{(6)} $$

Let $2y+1$ to be denoted as $z$. Which gives $$ x^3 = \dfrac{1}{4}\left(3z^2+1\right) \ \ \ \ \ \mathbf{(7)} $$ Which after rearranging gives the equation $$4x^3-3z^2 =1 \ \ \ \ \ \ \ \ \mathbf{(8)} $$ The question now boils down to checking integer solutions for a two variable equation rather than the $3$ variable equation from which we started. Clearly $(x,z) = (1,1)$ is an integer solution of this equation. I am ignoring these integer solutions because adding just $1$ unit cube to another integer side cube will never make a new cube of integer length. In fact it cannot even be a cube. (Imagine doing this in your mind.)

But I suspect that there are no more because of the following reason that the absolute difference $\left|4x^2-3x^3\right|$between the square and the cube term increases non-linearly, for e.g., say if we instead had the equation $$ 4x-3z=1 \ \ \ \ \ \mathbf{(9)}$$ The absolute difference $\left|4x-3y\right| $ increases linearly. Hence, there will be infinitely many integer solutions if there exists at least one integer solution.

But this is no way of proving statements. The last statement i.e. because of non-linear growth of the difference there will be no integer solutions is just out of my intuition. I have no formal way of proving there is no integer solution. How can approach proving such a thing? If proven rigorously that there will be no integer solutions this will be a complete proof of Fermat's Last Theorem for the case $n=3$ and for all cases $n$ being a multiple of $3$ (that is assuming there are no wrong assumptions which makes the proof wrong). If I am wrong anywhere in this attempted proof please correct me if possible.

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I can't comment since I don't have 50 reputation, but your question is basically the same as this post: Proving that this expression can never be a perfect cube

The techniques suggested there are to use some elliptic curve tech.

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  • $\begingroup$ Yes. That question is mine only. I did not get a proper response to that question so I thought instead of editing that question let me rephrase it in a new way and explain all the steps I went through so that people will engage in this one reading all the steps. $\endgroup$ – シャシュワト Mar 25 '18 at 20:08
  • $\begingroup$ @ShashwatSharan You may want to check it again, because there is now Sage output that has computed the integral points on an equivalent curve, which is a proper response in my book. $\endgroup$ – Valbord Mar 25 '18 at 20:09
  • $\begingroup$ Can you explain that response to me? I did not understand the output of that code. I have never used or heard of sage before. $\endgroup$ – シャシュワト Mar 25 '18 at 20:11
  • $\begingroup$ @ShashwatSharan Sage is an open source computer algebra system that can handle objects like elliptic curves quite well. What that little script did was construct the elliptic curve $y^2=x^3-432$ and then find the integral points on that curve, which were $(x,y)=(12,36)$. Since the reduction was in terms of $(m,w)$ and not $(x,y)$ the solution corresponding to your curve is $(1,1)$, which is what you wanted to show. $\endgroup$ – Valbord Mar 25 '18 at 20:17
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    $\begingroup$ @ShashwatSharan You are of course welcome to do the same computation by hand, but be aware that it can get quite long for general curves. Yours is nice and simple, so you shouldn't have a problem, but it truly is not worth the effort. Every professional mathematician in the world will accept the statement "Sage told me there was only one point" as a complete and rigorous proof, particularly if you include your code in an appendix. $\endgroup$ – Valbord Mar 25 '18 at 20:40

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