Recent content by barylwires

I should have written that, by hyp., gcd(a, b)=1 \Rightarrow gcd(a^n, b^n) = 1, then gcd(a^n, b^n) = b^n = 1. Does that do it? I'm referring specifically to these, particular, relatively prime a and b. Clearly, gcd(4^2, 6^2) = 4 ≠ 2 = gcd(4, 6) is a counter example of gcd(a^n, b^n) = gcd (a, b)...

Groan. Must have been thinking a^n|b^n \Rightarrowa|b, but writing something entirely different. I suppose sleep is everything they say it is.

Many thanks for all the good pushing. What I have is clunky at best, and probably flawed: n = (a/b)^n \Rightarrowb^n|a^n \Rightarrowgcd(a^n, b^n) = b^n = gcd(a, b) = 1 \Rightarrowb=1 \Rightarrown=a^n. Since n≥2, a>1, but n≤2^n \Rightarrow\Leftarrow

Thanks. My initial contradictory supposition was that n^(1/n) was rational. My biggest problem here (I think) is seeing where rational a/b must be integer a/1. Is that what the hint is supposed to help me see? That no integer x can have x or more factors, together with gcd(a, b)= gcd(a^n, b^n)...

I see that n has r<n prime factors, and that n|a^n. And a^n/n = b^n. I can't see the connection between this feature, the fact that a and b are co-prime, and a contradiction.

Why are primes dividing n on the RHS? Do you mean I should consider the division of n by primes? As is , the problem implies two divisions: n|a^n and b^n|a^n.

I don't think I'm using the hint in the following way; is it correct? n=a^n/b^n --> a^n=n•b^n --> b^n|a^n --> b|a --> n^(1/n) is an integer, but by hyp., n≥ 2 --> 2^(1/2) is an integer --><-- (relies, of course, on sqrt(2) known to be irrational)

1. For n ≥2, n^(1/n) is irrational. Hint provided: Use the fact that 2^n > n2. This is probably familiar to many. By contradiction, n = a^n/b^n --> a^n = n(b^n) --> n|a^n --> n|a Am I trying to force the same contradiction as with 2^1/2 is rational, that is, that a/b are not in lowest terms? Or...

That's beautiful! Thanks.

I mean, the left side is divisible by 24 by induction hypothesis. Why do I get the right side for free? Or, do I?

I see that 12(7^k + 5^k) is divisible by 24 when k=1, but why does that suffice for all k>1? Don't I have to prove that?

Good things happen. I think I can write: Assume aRb. Then 5|2a+3b iff 5|-(2a+3b) iff 5|-2a-3b iff 5|-3b-2a+5b+5a iff 5|2b+3a. Now aRb implies bRa, thus R is symmetrical. I'll sleep on that. Many thanks!

Sorry. I'm trying to show symmetry, not transitivity.

1. aRb on Z if 5|2a+3b 2. Since 5|2a+3b, 2a+3b=5m, so 2a=5m-3b. 3. Consider 3a+2b=2a+2b+a=5m-3b+2b+a=5m-b+a. I can't show that 5 divides that, but there is another way to wrangle that into a better form. I need help seeing how that works. Thanks.