To begin with, here we only consider the case of n3+(n+1)3=(n+2)3, that is the case of consecutive integers (x,y,z) in the well known equation. The general case x3+y3=z3 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 n3 as a sequence of consecutive odd numbers. We get the sequence by writing n3=n⋅n2. This means that we need n terms with an average value of n2 to represent the cube n3. This representation has a big advantage over other possible representations and it is that consecutive cubes n3, (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 23=2⋅22=3+5 and 33=3⋅32=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.
63+73=83.
We write down the representation of each sequence:
63=6⋅62=31+33+35+37+39+41, that is 6 terms with average value of 62=36.
We do the same for 73.
73=7⋅72=43+45+47+49+51+53+55, that is 7 terms with average value of 72=49.
When we concatenate the two sequences into one we will call lhs, we get:
lhs=31+33+35+37+39+41+43+45+47+49+51+53+55.
This sequence has (6+7)=13 terms with average (n2+(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⋅43=559=13⋅43.
The right hand side is rhs=83=8⋅82, so it will have 8 terms with an average of 64. rhs=57+59+61+63+65+67+69+71. So looking at the result, we see that lhs=13⋅43 cannot be equal to rhs=83. That is (2n+1)⋅(n2+(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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