
#1
Jun2004, 01:27 PM

P: 95

if a number is a cube and a square the only forms will be 9k or 9k+1. any suggestions as to how to vaidate this?? would the 8 cases work? from 9k to 9k+8? what does everyone else think?
yet another... 10 divides z if and only if (10,z) does not = 1. 



#2
Jun2004, 01:54 PM

P: 696





#3
Jun2004, 01:57 PM

P: 95

that really doesn't help much, i kind of thought of that. this is one of those proofs that need to show one way then the other. that is all i am coming up with.




#4
Jun2004, 01:58 PM

P: 696

inquisitive minds 



#5
Jun2004, 02:19 PM

P: 95

so if i have the left hand side saying that if you choose n = 1, then that says that (10,n) cannot = 1. with = to 2, it says the same thing, so it would work for all n except for 10, and that would givbe you 1. the right hand side would say that (10,n) not = 1. could i assume that it DOES = 1 and show a contradiction? would that be valid for this type of proof?




#6
Jun2004, 02:22 PM

P: 95

does anyone have any suggestions on the first question? would showing the 8 cases be the easiest way to p[rove this? i am thinking so, just square them and cube at the same time, or should i square them first THEN cube?




#7
Jun2004, 03:34 PM

Emeritus
Sci Advisor
PF Gold
P: 11,154

For a number, k, to be a square and a cube, it needs to be the 6th power of another number. This is evident from the prime factorization of k. So, we need to show that
[tex] k = n^{6m} \equiv 0 or 1 (mod 9) [/tex] So we need consider only the nine cases n=0,1,2,...,8 [tex]0^{6} = 0 \equiv 0 (mod 9)[/tex] [tex]1^{6} = 1 \equiv 1 (mod 9)[/tex] [tex]2^{6} = 64 \equiv 1 (mod 9)[/tex] [tex]3^{6} \equiv 6^{6m} \equiv 0 (mod 9)[/tex] [tex]4^{6} = 2^{12} \equiv 1 (mod 9)[/tex] [tex]5^{6} \equiv (4)^6 = 4^6 \equiv 1 (mod 9)[/tex] [tex]7^{6} \equiv (2)^6 = 2^6 \equiv 1 (mod 9)[/tex] and [tex]8^{6} \equiv (1)^6 = 1 (mod 9)[/tex] And of course, 0^m = 0 and 1^m = 1, so that completes the proof. 


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