inquisitive minds

1+1=1

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.

Muzza

Obviously if 10 divides z, then (10, z) is not 1 (it must be at least 10). But the converse is not true, consider z = 2...

1+1=1

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.

Muzza

But, uh, didn't I just give a counterexample to the other implication? And thus, it's false?

1+1=1

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?

1+1=1

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?

Gokul43201

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.