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https://news.ycombinator.com/item?id=12018221
... and previously submitted without discussion:
_DON'T DOWN VOTE JUST BECAUSE YOU CAN'T DO MATH_ The proof Fermat hinted to was about the difference between squares. All whole numbers taken to a power greater than two (n^3) can be represented as the difference between two whole squares (x^2 - y^2). These differences can then be shown as the sum of consecutive odd numbers: 2^3 = 3^2 - 1^2 = (1+3+5) - (1) = 8, 3^3 = 6^2 - 3^2 = (1+3+5+7+9+11) - (1+3+5) = 27, 4^3 = 1…
Sounds interesting, but not sure what you mean by: > you'll discover that there will always be a gap if you try and combine two odd number series together Can you elaborate?
This proof (or "plausibility argument") bugs me so much. Just because something thins out and becomes rare doesn't mean it doesn't exist. As n gets bigger, the probability of n being a perfect square gets smaller and smaller. In the limit, the probability is zero. Does this mean square numbers don't exist?
This isn't a proof.
Schwartz distributions, infintesimals; okay, fine, those turned out to be a weird trick that can be formalised. But sometimes their tricks are just plain wrong and this is one example of a trick that just is wrong and can't be formalised.
_DON'T DOWN VOTE JUST BECAUSE YOU CAN'T DO MATH_ The proof Fermat hinted to was about the difference between squares. All whole numbers taken to a power greater than two (n^3) can be represented as the difference between two whole squares (x^2 - y^2). These differences can then be shown as the sum of consecutive odd numbers: 2^3 = 3^2 - 1^2 = (1+3+5) - (1) = 8, 3^3 = 6^2 - 3^2 = (1+3+5+7+9+11) - (1+3+5) = 27, 4^3 = 1…
can you demonstrate?
_DON'T DOWN VOTE JUST BECAUSE YOU CAN'T DO MATH_ The proof Fermat hinted to was about the difference between squares. All whole numbers taken to a power greater than two (n^3) can be represented as the difference between two whole squares (x^2 - y^2). These differences can then be shown as the sum of consecutive odd numbers: 2^3 = 3^2 - 1^2 = (1+3+5) - (1) = 8, 3^3 = 6^2 - 3^2 = (1+3+5+7+9+11) - (1+3+5) = 27, 4^3 = 1…
Do you really believe that: (a) This constitutes a proof; (b) This is the "proof" that Fermat had; (c) Mathematicians missed this for over 350 year? I'm not quite sure exactly what you are claiming.
This proof (or "plausibility argument") bugs me so much. Just because something thins out and becomes rare doesn't mean it doesn't exist. As n gets bigger, the probability of n being a perfect square gets smaller and smaller. In the limit, the probability is zero. Does this mean square numbers don't exist?
Roughly, it goes as such:
1) the probability of N being a perfect square is proportional to 1/sqrt(N).
2) For any N_0 arbitrarily high, if you integrate from N_0 to infinity the expression (1/sqrt(N) dN), you get infinity.
3) The expression in 2) is the "Feynman equivalent" of the expected number of square numbers above N_0.
So Feynman's nonproof actually turns out to be true, despite it not being a proof in this case as well.
_DON'T DOWN VOTE JUST BECAUSE YOU CAN'T DO MATH_ The proof Fermat hinted to was about the difference between squares. All whole numbers taken to a power greater than two (n^3) can be represented as the difference between two whole squares (x^2 - y^2). These differences can then be shown as the sum of consecutive odd numbers: 2^3 = 3^2 - 1^2 = (1+3+5) - (1) = 8, 3^3 = 6^2 - 3^2 = (1+3+5+7+9+11) - (1+3+5) = 27, 4^3 = 1…
> It's not that hard people. Stop believing everything you're told about how "hard" something is. There are still many problems in physics and mathematics which are considered "hard" (e.g., dark energy, Riemann hypothesis, etc). Can we crack them by simply adopting your positive mindset?
_DON'T DOWN VOTE JUST BECAUSE YOU CAN'T DO MATH_ The proof Fermat hinted to was about the difference between squares. All whole numbers taken to a power greater than two (n^3) can be represented as the difference between two whole squares (x^2 - y^2). These differences can then be shown as the sum of consecutive odd numbers: 2^3 = 3^2 - 1^2 = (1+3+5) - (1) = 8, 3^3 = 6^2 - 3^2 = (1+3+5+7+9+11) - (1+3+5) = 27, 4^3 = 1…
> The same trick works for higher powers. can you demonstrate?
x and y will be a multiple of the base number.
_DON'T DOWN VOTE JUST BECAUSE YOU CAN'T DO MATH_ The proof Fermat hinted to was about the difference between squares. All whole numbers taken to a power greater than two (n^3) can be represented as the difference between two whole squares (x^2 - y^2). These differences can then be shown as the sum of consecutive odd numbers: 2^3 = 3^2 - 1^2 = (1+3+5) - (1) = 8, 3^3 = 6^2 - 3^2 = (1+3+5+7+9+11) - (1+3+5) = 27, 4^3 = 1…
Sounds interesting, but not sure what you mean by: > you'll discover that there will always be a gap if you try and combine two odd number series together Can you elaborate?
Earlier quoted context omitted.
Do you really believe that: (a) This constitutes a proof; (b) This is the "proof" that Fermat had; (c) Mathematicians missed this for over 350 year? I'm not quite sure exactly what you are claiming.
It's completely arrogant to assume that because it hasn't been solved by "better" people that I couldn't solve it.