true or false ? does solutions exists in natural numbers for n>2 ?
(1/p)^n + (1/q)^n = (1/r)^n
why not ? why yes ?
2006-11-02 02:03:16 · 10 answers · asked by ramesh the great 1 in Science & Mathematics ➔ Mathematics
Let z = 1/n, q =1/p and s=1/r
Then z^n + q^n = s^n
This is Fermat's theorem and it was recently proven to hold only for numbers of n = 1 or 2.
Therefore The answer is No.
2006-11-02 02:16:01 · answer #1 · answered by ironduke8159 7 · 0⤊ 2⤋
False.
If that were true ( n>2 ) that would be the same as writing
q^n + p^n = (p*q/r)^n
(by multiplying both members by p^n*q^n )
It has been demonstrated since 1995 (Fermat's Theorem)
that you cannot find p, q, z integers such that
q^n + p^n = (p*q/r)^n = z^n for n>2
Or, il you prefer to have all integers,
(rp)^n + (rq)^n = (pq)^n
2006-11-02 10:26:17 · answer #2 · answered by Duke_Neuro 2 · 0⤊ 0⤋
Multiply everything by (pqr)^n, you get
(qr)^n + (pr)^n = (pq)^n
where pq, pr and qr are integers. This contradicts Fermat's Last Theorem, so it can't happen.
Answers just above and below mine are incomplete because they still deal with fractions, not integers.
2006-11-02 10:22:48 · answer #3 · answered by Anonymous · 0⤊ 0⤋
Yes, it is true, only if "p" or "q" are not "0". It is a sum of fractions.
2006-11-02 10:13:40 · answer #4 · answered by Anonymous · 0⤊ 0⤋
why not? false
why yes? true
2006-11-02 10:12:43 · answer #5 · answered by Anonymous · 0⤊ 0⤋
true
2006-11-02 10:10:31 · answer #6 · answered by soutelzfinezt 1 · 0⤊ 0⤋
true
2006-11-02 10:05:22 · answer #7 · answered by Anonymous · 0⤊ 0⤋
true
2006-11-02 10:03:56 · answer #8 · answered by Anonymous · 0⤊ 0⤋
how am I supposed to know I'm not a mathematician
2006-11-02 10:04:29 · answer #9 · answered by Mary Smith 6 · 0⤊ 2⤋
OK
2006-11-02 10:04:29 · answer #10 · answered by prizefyter 5 · 0⤊ 0⤋