What do you have so far?

Results 1 to 6 of 6

- Mar 22nd 2010, 04:07 PM #1

- Joined
- Mar 2010
- Posts
- 36

- Mar 23rd 2010, 12:34 PM #2

- Mar 23rd 2010, 02:06 PM #3

- Joined
- Mar 2010
- Posts
- 36

Ok so far I have no idea how to use ord p but I have made a proof which could be along the right lines....though its long here i go....

By FTA, we can express x,y as a product of primes uniquely:

x = (p1)^s1 x (p2)^s2...(pn)^sn

y = (p1)^t1 x (p2)^t2...(pn)^tn

Then x^a = (p1)^as1 x (p2)^as2...(pn)^asn

and y^b = (p1)^bt1 x (p2)^bt2...(pn)^btn

Since x^a = y^b,

(p1)^as1 x (p2)^as2...(pn)^asn = (p1)^bt1 x (p2)^bt2...(pn)^btn

=> asi = bti for all 1<= i <= n

so a divides bti and b divides asi.

Since gcd(a,b)=1, by Euler's lemma, a divides ti and b divides si.

=> ti = azi and si = byi fo some integer zi, yi

=> x^a = (p1)^aby1 x (p2)^aby2...(pn)^abyn

y^b = (p1)^abz1 x (p2)^abz2...(pn)^abzn

Again since x^a = y^b,

(p1)^aby1 x (p2)^aby2...(pn)^abyn =(p1)^abz1 x (p2)^abz2..(pn)^abzn

=> yi = zi for all 1<= i<= n

So x^a = (p1)^abz1 x (p2)^abz2...(pn)^abzn = [(p1)^z1 x (p2)^z2...(pn)^zn]^ab

=> x = [(p1)^z1 x (p2)^z2...(pn)^zn]^b

Similarly y^b = [(p1)^z1 x (p2)^z2...(pn)^zn]^ab

=> y = [(p1)^z1 x (p2)^z2...(pn)^zn]^a

Let n be some integer s.t. n = [(p1)^z1 x (p2)^z2...(pn)^zn]

Then we get x = n^b and y = n^a.

But do I have to worry about signs?? What about ord p? Is it possible you can give an argument as to how you would go about it??

Thanks!

- Mar 25th 2010, 04:14 PM #4

- Joined
- Mar 2010
- Posts
- 36

- Mar 25th 2010, 06:19 PM #5

- Mar 26th 2010, 06:08 AM #6

- Joined
- Mar 2010
- Posts
- 36