very simple question but tricky one?

Question

show that we have to prove the equation a^(n) - b^(n) = x^(n) of sir fermats last theorem only for x=all odd primes & n=all odd primes & 4.

Prove that when x is even then it is impossible to write the above equation For n>2.

Answer 1

Hello Rajeash
Your additional question "
Prove that when x is even then it is impossible to write the above equation For n>2.
"
Isnt this what thye good o'l Sir Fermat already has stated even for odd x ?


a^(n) - b^(n) = x^(n)
suppose that n is not prime, thus n = p*q, with p a prime
then we can write
(a^q)^p -(b^q)^p = (x^q)^p
so we can limit our problem to n a prime. Fermat stated his statement for p>2 thus we only have to investigate the odd primes.I do not agrede with You Rashesj that we have to investigate 4 as a separate case, because 4 = 2^2 we can write the equation as (a^2)^2 - etc , Do you agree with this ?

You also say that we can assume that if we have a solution then x will be an odd prime. If a,b,x have a common factor then we can divide that factor out. So we can assumet that they dont have a common factor. If x is even then a and b should be odd.

Answer 2

Ok. Note that the equation a^(n) - b^(n) = x^(n) is not the way Fermat's last theorem is usually stated, but it is equivalent.

If n is an multiple of two numbers [in other words, n has factors], let one of them (p) be either a prime or equal to 4, and the other one (k) be equal to n/p. Then if a^(n) - b^(n) = x^(n) then it would follow that (a^k)^p - (b^k)^p = (x^k)^p. So if we can prove that there is no solution for p, then we've also proved there is no solution for n (equal to k*p), because any solution for n would allow us to derive a solution for p.

Any number greater than 2 is either 4, an odd prime, or divisible by 4 or an odd prime. This is because all numbers of the form 2^x (other than 2) are divisible by 4, and all other numbers greater than 2 are either divisible by an off prime or are themselves an odd prime.

Answer 3

this has solution if n = 2

so we have to test for n =4
now let n be odd
if n is prime it is covered
if it is not prime let n = pq where p is a prime

a^n-b^n = x^n
can be written as

a^pq-b^(pq) = c^pq

so (a^q)^p - (b^p)^q = (c^p)^q

same as x^p-y^p = z^p when x = a^q etc

so if we prove for odd prime it is OK

if n is even then = 2^k or has odd number as a factor

odd case is covered

similarly we can prove that for 4 if it is proved it is ok

but such a statement is not valid for x odd prime as a,b,x may not have a common factor

Answer 4

7^(1)-3^(1)=4^(1)

Answer 5

dunno