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It is impossible to write
a^(n) - b^(n) =c^(2)
for all natural values of a , b & c when n>2.

2006-10-29 00:26:56 · 10 answers · asked by rajesh bhowmick 2 in Science & Mathematics Mathematics

10 answers

Hi Dear Rajesh

well i don't think its a homework , it is really good question
i must confess i really don't know , but you can solve it
as you said a , b and c are all natural numbers & n > 2
so if we get
a = 2 , b = 3 & n= 3
( 2 ^ 3 ) - ( 3 ^ 3) = 8 - 27 = -19
and ' - 19 ' cant be equall ' c ^ 2 '
c ^ 2 is a number like 25 , 25 = ( 5 ^ 2) so c = 5
so it is not correct Equation.

If you found the correct answer , pls let me know as well.
Good Luck dear.

{ and dear turkeyphant what he wants to find is not what you said he wants to prove " a^(n) - b^(n) =c^(2) " but what you said is a^(n) - b^(n) =c^(n) .the are compeletly different from eachother. } ....

2006-10-29 00:45:29 · answer #1 · answered by sweetie 5 · 6 3

I've been thinking about this. I wonder if it is true - for instance, 2^3 - (-1)^3 = 3^2. -1 is not a natural number so this is not a counterexample, but I don't really see where negative numbers would change things too much. (if integers are allowed rather than just naturals then of course b = 0 or c = 0 gives lots of solutions).

It might help if you let us know how the problem was assigned - are you doing modular arithmatic, perhaps?

2006-10-29 17:48:36 · answer #2 · answered by sofarsogood 5 · 0 0

wrong
take a=1, b=2, n=3
a^n-b^n=1^3-2^3
=1-8
= -7
which can never be c^2, as c^2 is always positive, so cannot be -7

2006-10-29 11:28:13 · answer #3 · answered by asha s 1 · 0 0

For n=2, the equation holds true because it then become a pythagoras equation. But for n>2, there exist some combinations for which the equation holds true, but for this a & b must be prime and 'n' should be even.

2006-10-30 02:16:00 · answer #4 · answered by Napster 2 · 0 0

If you rearrange the equation to c^n + b^n = a^n you simply gave a formulation of Fermat's famous last theorem.

The 17th-century mathematician Pierre de Fermat wrote in 1637 in his copy of Claude-Gaspar Bachet's translation of the famous Arithmetica of Diophantus: "I have a truly marvelous proof of this proposition which this margin is too narrow to contain." (Original Latin: "Cuius rei demonstrationem mirabilem sane detexi. Hanc marginis exiguitas non caperet.") However, no correct proof was found for 357 years, until it was finally proven using very deep methods by Andrew Wiles in 1995 (after a failed attempt a year before).

If you are interested in the history and solution of Fermat's theorem, read Fermat's Last Theorem by Simon Singh.

(please ignore sweeties's answer - she is a retard who can't even rearrange a simple equation - how do you expect her to solve one of the hardest problems in mathematics?)

2006-10-29 07:43:44 · answer #5 · answered by turkeyphant 3 · 0 1

Dude,

Who does your teacher thinks he is giving you this kind of homework? Is he stupid or looking for a genius in math?

People worked for more than 300 years trying to solve that problem.

If you are interested in the problem go to this link:

http://www-groups.dcs.st-and.ac.uk/~history/HistTopics/Fermat's_last_theorem.html

There you could find solutions and links to the most updated ones.

2006-10-29 08:04:16 · answer #6 · answered by Dr. J. 6 · 0 1

if yu try hard enough for n even you may find phytagorian triplets with the desired property.

2006-10-29 15:29:40 · answer #7 · answered by gjmb1960 7 · 0 1

as you defined it, it must be wrong, since even one case that does not follow this equation is enough to reject the assumption that it is always right.

2006-10-29 07:41:16 · answer #8 · answered by Anonymous · 0 2

Ya do your own homework

2006-10-29 07:35:58 · answer #9 · answered by Anonymous · 0 1

hey thicko, go do your own homework!

2006-10-29 07:32:00 · answer #10 · answered by Anonymous · 0 1

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