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To begin with, here we only consider the case of $n^3 + (n+1)^3 = (n+2)^3$, that is the case of consecutive integers $(x,y,z)$ in the well known equation. The general case $x^3 + y^3 = z^3$ is much more difficult to prove (and the work is yet to be done with the method outlined below).

Here we use the integer representation of a cube $n^3$ as a sequence of consecutive odd numbers. We get the sequence by writing $n^3=n\cdot n^2$. This means that we need $n$ terms with an average value of $n^2$ to represent the cube $n^3$. This representation has a big advantage over other possible representations and it is that consecutive cubes $n^3$, $(n+1)^3$, and $(n+2)^3$ have a representation that ends when the next one starts (that is they do not share any term). For example $2^3=2\cdot 2^2= 3+5$ and $3^3=3\cdot 3^2=7+9+11$. This is a well known property of cubes when represented as a sequence of consecutive odd integers (https://en.wikipedia.org/wiki/Cube_(algebra)). This means that we can concatenate the left hand side of the equation to form a new sequence. And we need to figure out what integer this concatenated sequence represents. It was not really difficult to do that. So we gave an example and show why the left hand side cannot represent a cube whose value is the right hand side of the equation. $$6^3 + 7^3 = 8^3.$$ We write down the representation of each sequence: $$6^3 = 6\cdot 6^2 = 31 + 33 + 35 + 37 + 39 + 41,$$ that is 6 terms with average value of $6^2=36$. We do the same for $7^3$. $$7^3 = 7\cdot7^2 = 43 + 45 + 47 + 49 + 51 + 53 + 55,$$ that is 7 terms with average value of $7^2=49$. When we concatenate the two sequences into one we will call lhs, we get: $$\text{lhs} = 31 + 33 + 35 + 37 + 39 + 41 + 43 + 45 + 47 + 49 + 51 + 53 + 55.$$ This sequence has $(6+7)=13$ terms with average $(n^2 + (n+1)^2 +1)/2$. In our case, the lhs is therefore a sequence with 13 terms with average 43 which represents the integer $m=13\cdot 43=559=13\cdot43$.

The right hand side is $\text{rhs}=8^3=8\cdot8^2$, so it will have 8 terms with an average of $64$. $$\text{rhs} = 57 + 59 + 61 + 63 + 65 + 67 + 69 + 71.$$ So looking at the result, we see that $\text{lhs}=13\cdot 43$ cannot be equal to $\text{rhs}=8^3$. That is $(2n+1)\cdot (n^2 + (n+1)^2 +1)/2$ cannot be equal to $(n+2)^3$.

I am not sure if this method can be extended to the general case where $x,y$ and $z$ can take any value (not necessarily consecutive). But I think it's worth looking into it because of the simplicity of the method.

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  • 6
    $\begingroup$ https://en.wikipedia.org/wiki/Paragraph $\endgroup$ – arctic tern Oct 15 '16 at 19:38
  • $\begingroup$ Try to learn, and quick, how to use MathJax in this site, of which you've been a member for more than 4.5 years (!) and have already asked 9 questions and also answered 9 questionss. I mean, if you have some hope more than 1-2 people actually read your piece. $\endgroup$ – DonAntonio Oct 15 '16 at 19:41
  • $\begingroup$ What is the Question here? $\endgroup$ – hardmath Oct 15 '16 at 20:06
  • $\begingroup$ Considering that $n^3+(n+1)^3=2n^3+3n^2+3n+1=(n+2)^3=n^3+6n^2+12n+8$ is such a narrow, specific and uninteresting case any proof longer than a single line is probably not worth the trouble. $\endgroup$ – fleablood Oct 15 '16 at 20:21
  • $\begingroup$ @Hagen von Eitzen, thank you. $\endgroup$ – user25406 Oct 15 '16 at 20:34
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No.

The presumably shortest proof for the case you consider is this

Assume $$n^3+(n+1)^3=(n+2)^3. $$ Then $$ 0=n^3+(n+1)^3-(n+2)^3= n^3-3n^2-9n-7=(n^2-3n-9)\cdot n-7.$$ We conclude that $n\mid 7$, but neither $1^3+2^3=3^3$, nor $7^3+8^3=9^3$. $\square$

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  • $\begingroup$ I agree with you but we still haven't answered the question of the use of sequences of odd integers to represent integers to tackle the more general case. Is-it worth it? I suppose it is because we don't really know the answer. $\endgroup$ – user25406 Oct 15 '16 at 20:42
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$n^3 + (n+1)^2 = (n+2)^3 \implies $

$(m-1)^3 +m^3=(m+1)^3; m=n+1$

$2m^3-3m^2+3m-1=m^3+3m^2+3m+1$

$m^3=6m^2+2$

$m=6 + 2/m^2$ so $m^2|2$ and $m > 6$. So $m=1>6$. Clearly impossible.

Of course, the shortest proof would be "this is a single cubic equation with one variable and can easily be confirmed to have no integer solution".

This is simply far too specific a case to be of any significance.

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