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If r and s are relatively prime, then U(r) * U(s) = U(r*s). | If r and s are relatively prime, then U(r) * U(s) = U(r*s). | ||
| − | --[[User:Ysuo|Yu Suo]] | + | :--[[User:Ysuo|Yu Suo]] |
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| + | I discovered the same conjectur: U(r) * U(s) = U(r*s) | ||
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| + | :--[[User:Jzage|John Zage]] | ||
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| + | Obviously the conjecture they are leading you to is that U(r)*U(s)=U(r*s) but only if gcd(r*s)=1 | ||
| + | |||
| + | :--[[User:Ddesutte|Dane DeSutter]] | ||
Latest revision as of 19:00, 5 February 2009
For this problem, does anyone know what the new conjecture is supposed to be? I thought it might just be that the order of r multiplied by the order of s is NOT the order of rs, but I wasn't sure if there was another conjecture that could be made. --Clwarner 21:14, 3 February 2009 (UTC)
The conjecture I made was that $ \scriptstyle gcd(r,s)=1\ \leftrightarrow\ \mid U(r)\mid*\mid U(s)\mid=\mid U(rs)\mid $.
- --Nick Rupley 05:01, 4 February 2009 (UTC)
I made an equivalent conjecture: If r and s are relatively prime, then U(r) * U(s) = U(r*s).
- --Yu Suo
I discovered the same conjectur: U(r) * U(s) = U(r*s)
Obviously the conjecture they are leading you to is that U(r)*U(s)=U(r*s) but only if gcd(r*s)=1
