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Here is a basic geometric argument. Consider the sum 3^2 + 4^2 = 5^2. We can represent this as three "squares" of ones, such as:

1 1 1     1 1 1 1     1 1 1 1 1 
1 1 1  +  1 1 1 1  =  1 1 1 1 1 
1 1 1     1 1 1 1     1 1 1 1 1
          1 1 1 1     1 1 1 1 1 
                      1 1 1 1 1

Where each number is represented as the sum of ones in each square.

Now, consider the case of a^3 + b^3 = c^3. For illustrative purposes, let's assume that, for instance, 3^3 + 4^3 = 5^3. If this was true, it could be written in the form:

3 3 3   4 4 4 4    5 5 5 5 5 
3 3 3 + 4 4 4 4  = 5 5 5 5 5
3 3 3   4 4 4 4    5 5 5 5 5 
        4 4 4 4    5 5 5 5 5
                   5 5 5 5 5

In essence, we have our squares again, only the "units" are now equal to the length of a side.

Assumption -- all powers > 2 can be expressed this way, meaning essentially as squares but with the appropriate units. So, for instance, the number 3^5 can be expressed as:

27 27 27             9 9 9             3 3 3            1 1 1
27 27 27  and 3^4 =  9 9 9  and 3^3 =  3 3 3  and 3^2 = 1 1 1
27 27 27             9 9 9             3 3 3            1 1 1

So, then, for each triple (a,b,c), this assumption implies that if a solution to a^x + b^x = c^x exists, then a^2 + b^2 = c^2 must also be true. Since both of those things cannot be true, this shows that triples of higher powers cannot exist.

What I don't know is if this is sufficient to show that every possible triple of a higher power therefore cannot exist.

EDIT: This is way too "easy", I think, to be a proof of Fermat's Last Theorem by contradiction. I feel like I'm making an error someplace, namely in the assumption that if (a,b,c) is a "triple" in some higher power, then it also must be a Pythagorean triple.

EDIT2: I have removed the argument that a "triple" in higher powers implies that we can also do a^2/a + b^2/b = c^2/c and arrive at a+b=c, hence a contradiction.

EDIT3: I have changed the title to better reflect my actual question.

aplewe
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  • Your representation of $x^3$ as an $x\times x$ array of $x$s is fine, and I get what you are trying to do. However, $3^3+4^3\ne5^3$. In fact there are no positive integer solutions to $a^n+b^n=c^n$ for any $n>2$, this is known from Fermat's last theorem; it is a very famous result in number theory. – R. Burton Mar 11 '20 at 17:19
  • Yes, that's my point... If Fermat's Last Theorem wasn't true, then all sums of powers could be reduced to a pythagorean triple with the same roots. Or, at least, I think that's true and thus may demonstrate that Fermat's Last Theorem is true. – aplewe Mar 11 '20 at 17:23
  • Are you trying to write a proof of Fermat's Last Theorem? – R. Burton Mar 11 '20 at 17:24
  • Well, I'm trying to poke holes in this as a possible proof of Fermat's Last Theorem. I concocted this trying to come up with things that, perhaps, Fermat played with to come up with his theorem in the first place. – aplewe Mar 11 '20 at 17:25
  • It's too "easy", so I expect I'm making a fundamental error in my reasoning. – aplewe Mar 11 '20 at 17:26
  • You might want to consider explaining this in your question. – R. Burton Mar 11 '20 at 17:27
  • I don't want to be so bold as to claim this as a possible proof; I think the main error lies in assuming that all sums of powers, if such a thing were possible, could be reduced to sums of squares. So I'm trying to narrow the focus to that specific aspect. As if that's not true, then my "descent" to a proof by contradiction using sums of square roots also isn't true. I will edit the question and add this explanation to the body so my thinking and intent are clear. – aplewe Mar 11 '20 at 17:29
  • No, you are correct. For any $n$, $a^n+b^n=c^n\implies a^{n-2}a^2+b^{n-2}b^2=c^{n-2}c^2$, if you represent each square as a matrix of $1$s, then all you are doing is multiplying each matrix by the appropriate term. – R. Burton Mar 11 '20 at 17:32
  • Perhaps, then, it is the last step -- the assumption that a^x + b^x = c^x implies that a^2/a + b^2/b =c^2/c and therefore sqrt(a+b) = sqrt(a) + sqrt(b)? Although, this might not be necessary. – aplewe Mar 11 '20 at 17:36
  • I'm not sure how you got this last step. – R. Burton Mar 11 '20 at 17:38
  • If this was a proof, it would ridicule three centuries of the strongest mathematicians and the Field Medal would turn to chocolate. There is no particular relation between Pythagorean triples and higher degree triples (which do not exist). –  Mar 11 '20 at 17:45
  • I agree, hence this must be wrong somehow. Or, rather, it's sound but not complete/sufficient to "prove" anything. – aplewe Mar 11 '20 at 17:46
  • Your sentence "So, then, …" is nowhere justified. –  Mar 11 '20 at 17:47
  • I think the sqrt bit is not justified. The other question is if every case of a sum of equal powers implies the existence of a sum of squares. – aplewe Mar 11 '20 at 17:51
  • sqrt(a+b) != sqrt(a)+sqrt(b) – R. Burton Mar 11 '20 at 18:45

1 Answers1

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If $a^n+b^n=c^n$, then $\sqrt[n]{a^n+b^n}=c$ : true

$\sqrt[n]{a^n+b^n}=\sqrt[n]{a^n}+\sqrt[n]{b^n}$ : false

R. Burton
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