That’s true, although I actually think the main problems here were just the language barrier and the disruptive arrogance. It might otherwise have been reasonably eye-opening conversation for some bright but naive high school kid interested in math.
I think he means, for example,
x^4 = a^4 + b^4 + c^4
This would be an example of his equation
x^n=w1^n+w2^n+w3^n+ . . . wn-p^n
where n = 4 and p = 1.
Well, when n = 3 and p = 1, for example, that would produce x^3 = a^3 + b^3, which is unsolvable by Fermat. So is the claim that this is solvable for all n and p? For every n, there is a p for which this is solvable? This is unsolvable for all n and p? What, exactly?
Perhaps he’s “rediscovered” Euler’s conjecture.
That does look close. Interesting.
Anyway, back to aruvqan’s question - number theory (the field in which Fermat’s theorem exists) turns out to have practical uses. So do many other fields which might seem at first to be pretty abstract. For that matter, computer science grew out of the study of the foundations of mathematics (arguably a far more abstract field than number theory); you can’t tell what’s going to have practical applications and what’s not.
An older thread of mine:
http://boards.straightdope.com/sdmb/showthread.php?t=451945&highlight=fermat's+Pythagorean
.
I’m extremely sorry about the impression you guys received yesterday and I am very embarreased. What hamster said is what I meant. P is equal to one in certain instances such as when the power is 8. But is also naught in other instances such as a^3=x^3+y^3+z^3. I love maths and I am so glad to find people who share this passion. I would deeply appreciate it if you would help me prove this conjecture.
Does the diophantine expression
X^n=a^n +b^n+…+(N-1)^n +N^n
where all terms a through n have all integer values?
If that is what you’re trying to ask, that is seriously deep…even deeper than Fermat.
Wait, what? This is still confusing. Slow down, take 10-15 minutes, and accurately describe your conjecture. You state p = 1 in some circumstances but p is not 1 in other circumstances. Clearly state those circumstances. Show examples, if you have to. It does you no good if nobody understands the basic statement.
From what I can tell, your conjecture is something along the lines of “This stuff is true in some cases but not in other cases and categorization of the cases is not forthcoming”. That’s pretty darned vague.
Some help in actually proving the statement might be forthcoming if anybody could figure out what the conjecture actually states.
Also, putting on an arrogant facade and making nigh incomprehensible statements is not the best way to request aid.
No that isn’t what Im saying I’m saying that x^n=a1^n +a2^n + a3^n . . . an-p^n
n is a whole number. n-p cannot equal 2 unless p=0 and n=2.
It is more like Euler’s equation except it accepts that p>0.
What about this equation? Are you saying, for every n, there is a p, an x, and a choice of a_1 through a_{n-p} which makes this equation true? (In other words, “For every n, there is a sum of less than n many nth powers which is itself an nth power”)
Are there any restriction on how large p can be chosen to be? Because one can simply make n - p = 1, and then take x and a1 to be equal, to trivially solve such an equation.
Try just stating your claim in words instead of not fully explained notation; it may be easier for us to understand what you are trying to say that way.
n-p cannot equal 2 or 1 unless p=0. Then for every n, there is a sum of less or equal value to n many nth powers which is itself an nth power. That s important less or equal to. I don’t know if p>1 but it does sometimes equal 1.
Can you help me prove it or is my explaining really that bad?
I’m having trouble following exactly what you’re claiming. Can you talk a bit about how your claim differs from Euler’s sum of powers conjecture?
If I understand Euler’s conjecture correctly he said to the nth terms. I’m saying to the nth terms MINUS p. Such that n-p =2 if and only if p=0. n-p=1 if and only if p=0. This is to not disobey fermat.
What is p?
p is how many nth terms there are are less than n
That would imply that p = n-1? So n-p is always exactly 1?
no what if the difference is zero? Then p=0
Do you think you can prove that this?