Gareth, Well done! Thanks! R. On Sat, 19 Apr 2008, Gareth McCaughan wrote:
On Friday 18 April 2008, Richard Guy wrote:
This isn't much fun, but some of you with access to algebra manipulation systems may be able to help. Etienne Garnier has pointed out what must be a mistake in D13 of UPINT. I'm not able to access the relevant paper or even a review thereof, so could someone check what value of z (if any) should replace the one given below in the solution of the equation
x1^x1 * x2^x2 * ... * xk^xk = z^z ? ... x_1 = k^{k^n(k^{n+1}-2n-k)+2n}(k^n-1)^{2(k^n-1)} x_2 = k^{k^n(k^{n+1}-2n-k)}(k^n-1)^{2(k^n-1)+2} x_3 = ... = x_k = k^{k^n(k^{n+1}-2n-k)+n}(k^n-1)^{2(k^n-1)+1} z = k^{k^n(k^{n+1}-2n-k)+n+1} ???????
(Plodding pedestrian solution follows. I have no computer algebra system worth the name, but this isn't really hard enough to require one.)
Simplify stepwise:
K := k^n-1
x_1 = k^{k^n(k^{n+1}-2n-k)+2n} K^{2K} x_2 = k^{k^n(k^{n+1}-2n-k)} K^{2K+2} x_3 = ... = x_k = k^{k^n(k^{n+1}-2n-k)+n} K^{2K+1}
a := k^n(k^{n+1}-2n-k)
x_1 = k^{a+2n} K^{2K} x_2 = k^{a} K^{2K+2} x_3 = ... = x_k = k^{a+n} K^{2K+1}
b := k^a K^{2K}
x_1 = b k^{2n} x_2 = b K^2 x_3 = ... = x_k = b k^n K
L := k^n
x_1 = b L^2 x_2 = b K^2 x_3 = ... = x_k = b K L
so now our LHS is
(bKK)^(bKK) (bKL)^[(k-2)bKL] (bLL)^(bLL) = b^(bKK+(k-2)bKL+bLL) K^(2bKK+(k-2)bKL) L^(2LL+(k-2)bKL) = b^(kbKL+b(K-L)^2) K^(kbKL+2bK(K-L)) L^(kbKL+2bL(L-K)) { and since L-K=1 ...} = b^(kbKL+b) K^(kbKL-2bK) L^(kbKL+2bL)
and it's time to split b up again:
= k^(kabKL+ab) K^(2kbKKL+2bK) K^(kbKL-2bK) L^(kbKL+2bL) = k^(kabKL+ab) K^(2kbKKL+kbKL) L^(kbKL+2bL)
and L, too:
= k^(kabKL+ab+nkbKL+2nbL) K^(2kbKKL+kbKL)
This will equal (k^p K^q)^(k^p K^q) provided kabKL+ab+nkbKL+2nbL = p k^p K^q 2kbKKL+kbKL = q k^p K^q
and it's about time we pulled out some factors from what are now our LHSes: we need b(kaKL+a+nkKL+2nL) = p k^p K^q bkKL(2K+1) = q k^p K^q
so suppose q = 2K+1; then we need (from the second of those equations) bkKL = k^p K^q k^{a+n+1} K^{2K+1} = k^p K^q or in other words p = a+n+1.
Does this make the first equation work? Let's see: we want b(kaKL+a+nkKL+2nL) = (a+n+1) k^{a+n+1} K^{2K+1} k^a K^{2K} (kaKL+a+nkKL+2nL) = (a+n+1) k^{a+n+1} K^{2K+1} (kaKL+a+nkKL+2nL) = (a+n+1) k^{n+1} K
Well, a = L(kL-2n-k) so substitute that in: we want (kaKL+L(kL-2n-k)+nkKL+2nL) = (a+n+1) k^{n+1} K (kaK+kL-2n-k+nkK+2n) = (a+n+1) k K (kaK+kL-k+nkK) = (a+n+1) k K (aK+L-1+nK) = (a+n+1) K
and remember that L-1=K so that's (aK+K+nK) = (a+n+1) K
so indeed everything works.
Conclusion: p = a+n+1, q = 2K+1, so (unravelling our abbreviations)
z = k^p K^q = k^p (k^n-1)^q = k^{a+n+1} (k^n-1)^{2(2^n-1)+1} = k^{k^n(k^{n+1}-2n-k)+n+1} (k^n-1)^{2(2^n-1)+1}
which differs from the wrong formula only by the inclusion of the second factor.
[the exponent of k in z is less than that of x_1 and there is no factor k^n-1 at all]
The first of those isn't a problem; it's compensated for by the second factor.
-- g
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