Thanks.
(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.
Seriously, can you walk us through the steps of why x^5+y^5=z^5 has no (non-trivial) solutions?
And to be fair, there are cases where, say, undergrads have proven significant results that had been outstanding for a long time. Proving that prime recognition is in P is one such case. but in that case they published a complete, clear paper. In this case I can't really see what you're saying you've done, or why it's true, which is why a walk-through of the case n=5 would be so helpful.
Thanks.
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For anyone interested, this is what the comment used to say ...
The series of odd numbers must be consecutive and they simply are not when you add two different series together for all powers greater than two. It's okay. You'll get it.
1) a whole number, n, taken
to a power greater than two
2) can be represented as a
consecutive series of odd
numbers
3) where there will always be
a gap in the series between
consecutive base numbers,
for all p and p+1
4) therefore there will be
a gap for all p and p+n,
n>=1 combinationsOh, you're that guy.
~~Lemma 1~~:
z^n can be written as a difference of squares:
z^n = x^2 - y^2 = (x+y)(x-y), leading to the system of equations
x + y = z^(n-1)
x - y = z
which may be solved for x and y:
x = (z^(n-1) + z)/2
y = x - z
QED.
~~Lemma 2~~:
Any number squared can be written as a sum of sequantial odd numbers starting at 1.
(n)^2 = (n-1)^2 + (2(n-1)+1) = \sum_{k=0}^{n-1}(2k + 1)
QED.
~~~Fermat's Last Theorem~~~:
z^n = x^n + y^n has no solutions for n>2, and positive z, x, y.
Without loss of generality, assume a > b, then rewrite z^n as a sum of sequential odd numbers starting at b:
z^n = a^2 - b^2 = \sum_{k=b}^{a-1}(2k+1)
x^n and y^n can similarly be written as a sum of sequenatal odd numbers:
x^n = c^2 - d^2 = \sum_{k=d}^{c-1}(2k+1)
y^n = e^2 - f^2 = \sum{k=f}^{e-1}(2k+1)
By substitution to the Theorem's equation:
\sum_{k=b}^{a-1}(2k+1) = \sum_{k=d}^{c-1}(2k+1) + \sum_{f}^{e-1}(2k+1)
Which is true if and only if there are no gaps in the bounds of summation on the right side, so d = b, c = f, and e = a. But then
a^2 - b^2 = (f^2 + (-a^2))^n + (b^2 + (-f^2))^n
and this is certainly not true by the binomial theorem. We've reached a contradiction so QED.
^^ This is where I'm stuck. I don't actually know why that would not be true by the binomial theorem? It seems like simple expansion should do it.
> you'll discover that there will always be a gap if you try and combine two odd number series together
Can you elaborate?
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?
What does work though is this: advanced formal education in a topic. Once you have that you can start thinking on how to solve some simple open problems. And if you are lucky and turn out to be extremely smart, you may be able to tackle more challenging problems. Some amount of self confidence may also you to keep going but doesn't make you a genius overnight.
Simply going to a mindset where things are 'not hard' is closer to delusion than it is to anything else.
In academia we get often emails from people who solved quantum gravity (e.g. using fire), show us how einstein is wrong (e.g. using a pendelum), etc. I'm pretty sure they also convinced themselves to "Stop believing everything they're told about how "hard" something is"
It was pretty frustrating. He was too nice a guy to tell them off or even cut them off quickly.
My advice to any crackpots who are really sure they're actually geniuses: Get into the stock market (with a SMALL investment). If you're as smart as you think you are, you can find an angle and turn $100 into $1,000,000 or more, and then if anything it'll be GOOD that nobody ever believed in you. I've run across arbitrage opportunities that would have made me fiendishly rich if I'd noticed them sooner myself, believe it or not. Just be careful and don't mess with box spreads.
If you're asserting something that's likely to encounter resistance, it's worth being clear and careful.
can you demonstrate?
x and y will be a multiple of the base number.
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:
When you examine the odd number series that results from each base, you'll discover that there will always be a gap if you try and combine two odd number series together, which explains Fermat's little joke about margins. The same trick works for higher powers.It's not that hard people. Stop believing everything you're told about how "hard" something is.
HINT: The number of odd numbers in the series exactly matches the starting square base number