1. ## ord p

Let a,b be positive integers with gcd(a,b)=1.
Suppose x^a = y^b where x,y are integers.

Show that x = n^b and y = n^a for some integer n, using the properties of ord p .

2. What do you have so far?

3. 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!

4. Sorry to pester you, but any comments on what I have done? Thanks

5. Originally Posted by vinnie100
Sorry to pester you, but any comments on what I have done? Thanks
This arguement looks fine to me. I don't know how you would let order modulo p come into play...

For the sign, I would say
x = (-1)^c x (p1)^s1 x (p2)^s2...(pn)^sn
y = (-1)^d x (p1)^t1 x (p2)^t2...(pn)^tn