| To start with, define r = p/q, such that p and q are positive integers,
and p!|q. We are looking for all pairs (p,q) such that s = (p/q)^(q/(p-q))
is rational.
Since p/q is rational, obviously s will be rational when the exponent
is an integer. This occurs when (p-q)|q.
The other possibility is that the exponent is rational but not
integral. Let us assume that this is the case for some (p,q). Since
the (p-q)th root of both p and q are integral (this might be an
unfounded leap, I'm not sure), we can write p=a^n and q=b^n, such that
(p-q)|n, or (a^n - b^n)|n. Since a and b are integers, the smallest
that |a^n - b^n| for any given n can be is 2^n - 1. It is thus
possible for s to be rational only when (2^n - 1) <= n. This is
impossible for any n>1, which gives us a contradiction. Therefore
there exists no such pair (p,q) which are powers greater than 1.
So the only (p,q) that satisfy the condition are those for which
(p-q)|q. (Assuming, that is, my leap -- that the only way the (p-q)th
root of p and q can be commensurate is if they are both integral.
True?)
Matt
|
| re .1
>p!|q
I find this notation confusing since I read it as "p factorial divides q". I
worked out eventually that you mean p does not divide q.
re .0
Answer: r must be of form (q+1)/q or q/(q+1).
Proof:
Let r = p/q where (p,q) = 1
The expression is (p/q)^(q/(p-q)).
Consider first the case where p > q so that p-q > 0.
We want to find (p-q,q). Let (p-q,q) = t.
t|q and t|p-q => t|p
Since (p,q) = 1 this implies that t = 1 i.e. (p-q,q) = 1.
Consider then the two cases that p-q = 1 and p-q > 1.
If p-q = 1 (this is the "integer" case mentioned in .1) then
p = q+1 so r is of the form (q+1)/q.
Before considering the case that p-q > 1 we establish a few lemmas.
Lemma 1
-------
Where a and b are positive integers, a|b and b|a => a = b
Proof:
From a|b, let b = ak and from b|a, let a = bm
=> b/a = k and b/a = 1/m
=> k = 1/m
=> mk = 1
=> m = 1 (and k = 1)
=> a = b
Lemma 2
-------
For positive integers, a,b,d,x,y with (a,b) = 1 and (x,y) = 1
(a/b)^(1/d) = x/y => a = x^d and b = y^d
[This is the "leap" mentioned in .1.]
Proof:
(a/b)^(1/d) = x/y
=> a/b = (x/y)^d
=> ay^d = bx^d
=> a|x^d and b|y^d and x^d|a and y^d|b (using the gcd preconditions)
=> a = x^d and b = y^d (by Lemma 1)
Lemma 3
-------
For u a rational number and c, d such that (c,d) = 1,
u^(c/d) is rational => u^(1/d) is rational
Proof:
Let u = g/h and u^(c/d) = j/k where (g,h) = 1 and (j,k) = 1
=> g^c k^d = h^c j^d
=> g^c = j^d and h^c = k^d (proceding as in Lemma 2)
Now let t be an arbitrary prime divisor of g (so clearly t | j).
Let n(z) denote the number of times that t divides z, then
n(g)*c = n(j)*d
Hence, since (c,d) = 1, d | n(g)
=> t^d | g
=> g = j'^d for some j'
(perhaps the case g = 1 should have been covered specially but the conclusion
is obviously true in that case)
Similarly for h, h = k'^d for some k', thus
g/h = j'^d/k'^d
=> (g/h)^(1/d) = j'/k'
Hence u^(1/d) is rational
Now back to p-q > 1.
In that case, and given (p-q,q) = 1, we can see that
(p/q)^(q/(p-q)) is rational => (p/q)^(1/(p-q)) is rational, by Lemma 3.
Let d = p-q and apply Lemma 2 i.e. p = x^d and q = y^d.
Hence d = p-q = x^d - y^d
Noting that p-q > 1 => d >= 2 and x > y i.e. x-y >= 1,
we can see that
d = x^d - y^d
has no solutions by putting w = x-y i.e. x=y+w and expanding via the binomial
theorem giving
d = d*y^(d-1)*w + C(d,2)*y^(d-2)*w^2 + ...
Looking carefully at the terms shows that the first term is >= d and the second
term is >= 1 (in fact > 1). Consequently the RHS is strictly greater than d and
thus there can be no solutions to d = x^d - y^d.
Thus for p > q i.e. r > 1 the only solutions are of the form (q+1)/q.
Similarly considering the case p < q i.e. r < 1 gives that the only solutions
are of the form p/(p+1). [Did you see my hands wave?]
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