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 ? [the exponent of k in z is less than that of x_1 and there is no factor k^n-1 at all] Many thanks in anticipation. R. Claude Anderson conjectured that the equation $w^wx^xy^y=z^z$ has no solutions with $1<w<x<y<z$, but Chao Ko & Sun Qi had earlier found an infinity of counterexamples to a generalization of the conjecture to any number of variables: 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} ??????? Chao Ko & Sun Qi, On the equation $\prod^k_{i=1}x_i^{x_i}$, {\it J.\ Sichuan Univ.}, {\bf2}(1964) 5--9. [I may even have got the title of the paper wrong, since an ``= z^z'' seems to be missing.]
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
On 4/19/08, Gareth McCaughan <gareth.mccaughan@pobox.com> wrote:
... (Plodding pedestrian solution follows. I have no computer algebra system worth the name, but this isn't really hard enough to require one.) ...
Hm. Perhaps Richard will call me to heel if I'm jumping to conclusions: Chao & Sun apparently managed to stretch this into a 5-page paper. Funnily enough, the worldly mathematician who first introduced (also in the early sixties) to my tender young mind the shocking concept of a paper factory was also Chinese. The abysmally non-PC hypothesis suggests itself that our oriental colleagues might have been ahead of the pack, at least in this respect. [Come to think of it, I could have done without them inventing fireworks as well ...] Anyhow, impressively wrangled, Gareth! Fred Lunnon
On Saturday 19 April 2008, Fred lunnon wrote:
On 4/19/08, Gareth McCaughan <gareth.mccaughan@pobox.com> wrote:
... (Plodding pedestrian solution follows. I have no computer algebra system worth the name, but this isn't really hard enough to require one.) ...
Hm. Perhaps Richard will call me to heel if I'm jumping to conclusions: Chao & Sun apparently managed to stretch this into a 5-page paper.
I assume Richard doesn't have the paper, else he wouldn't have needed to ask math-fun's help. But maybe Chao and Sun explained how they found the result, or discussed further generalizations, or something. -- g
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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This might actually be a good exercise for a computer algebra system. -- Rich ________________________________________ From: math-fun-bounces@mailman.xmission.com [math-fun-bounces@mailman.xmission.com] On Behalf Of Gareth McCaughan [gareth.mccaughan@pobox.com] Sent: Friday, April 18, 2008 6:07 PM To: math-fun@mailman.xmission.com Subject: Re: [math-fun] Help! 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 _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
participants (4)
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Fred lunnon -
Gareth McCaughan -
Richard Guy -
Schroeppel, Richard