From: rusin@vesuvius.math.niu.edu (Dave Rusin) Newsgroups: sci.math Subject: Re: need help with this equation... Date: 24 Aug 1996 07:58:37 GMT In article <4vg7n8$jld@math.mps.ohio-state.edu>, John Lame wrote: > >This reminds me of something that I noticed some time ago but never >followed through on. While looking at the number of ways that a >given integer n may be represented as a sum x^2 + y^2 + z^2 with >x,y,z natural numbers such that 0 < x <= y <= z, I conjectured that: > >If n is a perfect cube having at least one such representation, then n >has more than one such representation. (e.g. 8 is not the sum of 3 >positive squares, 27 = 1^2 + 1^2 + 5^2 = 3^2 + 3^2 + 3^2, 64 has no >such representation, 125 = 3^2 + 4^2 + 10^2 = 5^2 + 6^2 +8^2, etc.) I fail to see what would prompt you to pick _cubes_ to represent as sums of _squares_, but, hey, whatever floats your boat. I once posted rather a long answer to a related question on sums of three squares. It sits with some related files at 95/three.sq [That's an updated URL -- djr] It seems you can use the arguments there to confirm your conjecture. (I hedge my answer only because I don't want to consider what it takes to ensure each of the summands is nonzero, as you required above). Here's the rough idea. 1. n^3 is a sum of 3 squares iff n is. (Proof: see whether n is of the form 4^t(8s-1). ) 2. n is a sum of four squares. (Proof: every integer is). OK, so if n^3 is a sum of 3 squares, find integers a b c d e f g so that n = a^2 + b^2 + c^2 = d^2 + e^2 + f^2 + g^2. Then n^3 = (na)^2 + (nb)^2 + (nc)^2 meaning that (n^3) is the norm of the quaternion x = (na)i + (nb)j + (nc)k. Let y be the quaternion y = d + ei + fj + gk. Then the conjugate y^(-1) x y is easily seen to have no real part (just like x), and its norm must equal norm(x) = (n^3), so that (n^3) has again been expressed as a sum of three squares. You have to be careful in your choice of y, in general, since this conjugate will have _rational_ rather than integral coordinates. But it may be rewritten (y-bar) x y / norm(y) = (y-bar) (x/norm(y)) y, which clearly has integral coefficients if the norm of y divides the coordinates of x. This was arranged by construction. You can see there's quite a bit of latitude here (e.g. it suffices to have norm(y) merely be a _divisor_ of n.) Example: n = 19. We may write n = 3^2 + 3^2 + 1^2 (and in fact that's the only description of 19 as a sum of three integral squares, up to changes of sign and of order). This of course gives 19^3 = 57^2 + 57^2 + 19^2. To get another representation, we write n as a sum of four squares; this time we have several options 19 = 4^2 + 1^2 + 1^2 + 1^2 = 3^2 + 3^2 + 1^2 + 0^2 and of course the permutations and sign changes. If I compute correctly I get for example (4 + i + j + k)^(-1) (57i + 57j + 19k) (4 + i + j + k) =(69 i + 37 j + 27 k) and indeed 19^3 = 69^2 + 37^2 + 27^2. An alternative proof runs like this: write n = n1^2 * n2 with n2 squarefree. Then n^3 has a representation as a sum of three squares iff n2 does. At the very least, we have among all representations of n^3 those of the form n^3 = (n*n1*a)^2 + (n*n1*b)^2 + (n*n1*c)^2 for every representation n2 = a^2 + b^2 + c^2. So then the question is, how many squarefree integers have a unique representation as a sum of three squares. The answer is, none except for those nine n2 for which the quadratic field Q(sqrt(-n2)) has class number 1, namely, n2=1, 2, 3, 7, 11, 19, 43, 67, 163 (n2=7 actually admits no 3-square representation.) In each of these cases, we find more than one representation for n2^3 as was done for 19, above. dave PS -- Is that Lam\'e like the 19th century FLT-solver? ============================================================================== From: ferriom@woods.uml.edu Newsgroups: sci.math Subject: Re: need help with this equation... Date: 26 Aug 96 12:03:31 -0500 Man Huu Nguyen wrote: > Would someone help me out? Let's say you have this equation: > x*x + y*y + z*z = m*m where x,y, and z are all integers, are >there more than one set of x,y,z where the sums will yield an integer, m, >also? Or put is another way, if you have a rectangular box, what >dimension should it be in order for the diagonal of the box to have an >integer value? I have come up with only on dimension so far (read further >to find out), but I don't know if there are others. Any thought would be >nice to hear... To solve x^2 + y^2 + z^2 = t^2 in integers, let x = m^2 - n^2 - p^2 + q^2 y = 2*m*n - 2*p*q z = 2*m*p + 2*n*q t = m^2 + n^2 + p^2 + q^2 Example: Letting m = 4, n = 3, p = 2, q = 1 gives 4^2 + 20^2 + 22^2 = 30^2, or just 2^2 + 10^2 + 11^2 = 15^2. -- Michael