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Topic: n = (x^2+y^2+z^2)/(xyz) (Read 8302 times) 

wonderful
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n = (x^2+y^2+z^2)/(xyz)
« on: Jun 19^{th}, 2008, 1:13am » 
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Find all positive integer number n satisfying the above equation given x, y, z are also positive integers. Have A Great Day!


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Hippo
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Re: n = (x^2+y^2+z^2)/(xyz)
« Reply #1 on: Jun 19^{th}, 2008, 5:30am » 
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There must be a trick ... I have obtained n=3 for several relative prime solutions (mod each of them 1 must be a square ...). Multiplying x,y,z by k the n is divided by k ... so n=1 is solution, too.

« Last Edit: Jun 19^{th}, 2008, 5:31am by Hippo » 
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denis
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Re: n = (x^2+y^2+z^2)/(xyz)
« Reply #2 on: Jun 19^{th}, 2008, 6:56pm » 
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Without loss of generality, let x<=y<=z Since n=(x^{2}+y^{2}+z^{2})/xyz Then 3x^{2}/z^{3}<=n<=3z^{2}/x^{3} with n integer (1) Case 1: x=1 clearly, x=1, z=1 (y=1) yields 3<=n<=3 > n=3 Case 2: x=3 Clearly x=3, z=3 (y=3) yields 1<=n<=1 > n=1 Now here is where I get stuck... I can't show for general case where x>3

« Last Edit: Jun 19^{th}, 2008, 10:59pm by denis » 
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Eigenray
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Re: n = (x^2+y^2+z^2)/(xyz)
« Reply #3 on: Jun 20^{th}, 2008, 5:00pm » 
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x^{2} + y^{2} + z^{2} = nxyz 1) As this equation is quadratic in each variable, if (x,y,z) is a solution, then so is (x,y,z'), where z' = nxyz = (x^{2}+y^{2})/z. 2) Let (x,y,z) be a solution with, WLOG, x <= y <= z, and z minimal. We therefore have that z^{2} <= x^{2}+y^{2}. 3) Now nxy^{2} <= nxyz = x^{2}+y^{2}+z^{2} <= 2(x^{2}+y^{2}) <= 4y^{2}, and at least one inequality above must be strict, so nx < 4. 4) The form y^{2}  (nx)yz + z^{2} can take on negative values only when nx > 2. Therefore nx=3, so n is either 1 or 3. In fact the minimal solution for n=1 is (3,3,3), and that for n=3 is (1,1,1), and we can show that any other solution can be obtained from one of these by repeatedly applying (1).

« Last Edit: Jun 20^{th}, 2008, 7:50pm by Eigenray » 
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wonderful
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Re: n = (x^2+y^2+z^2)/(xyz)
« Reply #4 on: Jun 20^{th}, 2008, 6:03pm » 
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Excellent Eigenray! Have A Great Day!


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Aryabhatta
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Re: n = (x^2+y^2+z^2)/(xyz)
« Reply #5 on: Jun 20^{th}, 2008, 7:10pm » 
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Yes. Nice one! I think this should be in medium though...


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Eigenray
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Re: n = (x^2+y^2+z^2)/(xyz)
« Reply #6 on: Jun 21^{st}, 2008, 12:03am » 
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To continue a bit: We can show that (3,3,3) is the unique minimal solution for n=1, and the rule (x,y,z) > (x,y,3xyz) preserves gcds. It follows that any solution (x,y,z) for n=1 has gcd(x,y,z)=3, so (x/3, y/3, z/3) is a solution with n=3. And conversely, of course, so there is a bijection between solutions for n=1 and those for n=3. Let us stick to n=3 now. Then any solution has x,y,z pairwise relatively prime. Connect two solutions if they differ in only one value. Since a minimal solution is unique, the set of all solutions forms a connected graph. Since the equation is quadratic in each variable, a solution (x,y,z) is only connected to (x,y,3xyz), (x,3zxy,z), and (3yzx,y,z). Note these are again all positive since, say, zz' = x^{2}+y^{2} > 0. Also note that if, say, z = z', then 3xy=2z so by relative primality, we'd have xy=2, z=3, which is not a solution. So each solution, as an ordered triple, is connected to exactly 3 others. Now suppose (x,y,z) is a solution with x,y <= z. Then x' = 3yz  x = z(3y1) + (zx) > z, and similarly y' > z. That is, each time we change a value other than the largest, it becomes the new largest. So if we repeatedly change values other than the one we just changed, the maximum will keep increasing, so there can be no loops. It follows that the set of all solutions forms an infinite 3regular tree. Furthermore, suppose, say, x=y. Write x=y=du, z=dv, where (u,v)=1. Then u^{2}(3dv2) = v^{2}. Since v is relatively prime to u, v3dv2, so v2. If v=1, then u=1, d=1, giving the solution (x,y,z)=(1,1,1). If v=2, u=1, d=1, giving the solution (1,1,2). So if we only look at solutions with x <= y <= z, we get an infinite binary tree, except for the two solutions (1,1,1) and (1,1,2) at the root. The equation is apparently known as Markov's equation, and the values that appear in solutions are called Markov numbers (see also here). By the way, this labelling of the regions around the tree of solutions reminds me of topographs from Conway's lovely little book "The Sensual (Quadratic) Form".

« Last Edit: Jun 21^{st}, 2008, 12:12am by Eigenray » 
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Eigenray
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Re: n = (x^2+y^2+z^2)/(xyz)
« Reply #7 on: Jun 21^{st}, 2008, 12:47am » 
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We can generalize to more variables too: a^{2} + ... + z^{2} = ma...z, where there are k variables (Hurwitz equation). The case k=1 is trivial. For k=2, we get m=2, x=y as the only solutions. Now fix k>2. Suppose a <= b <= ... <= z, and that the solution is minimal, so that z^{2} <= a^{2}+...+y^{2}. Then ma...xy^{2} <= ma...z = a^{2}+...+z^{2} <= 2(a^{2}+...+y^{2}) <= 2(k1)y^{2}, so ma...x <= 2(k1). This gives finitely many possible values for m,a,...,x. Suppose these are fixed, and let S = a^{2}+...x^{2}, P = ma...x. Then Py^{2} <= Pyz = S + y^{2} + z^{2} <= 2S + 2y^{2}. But P>2, or else y^{2}  Pyz + z^{2} >= 0, so this gives an upper bound on y too. Now for each of the finitely many values m,a,...,y, we can see if z is an integer. Thus, given k>2, we can effectively determine all possible values for m, and find one solution from each of the finitely many connected components. Moreover, suppose m >= k. Then a...x <= 2(k1)/m < 2, so a=...=x=1. As above, y^{2} <= 2S/(P2) = 2(k2)/(m2) <= 2, so y=1 as well. Finally, mz = (m1) + z^{2} <= 2(m1), and z=1 too, making m=k. So there is only one component with m>=k, namely the one with m=k, containing (1,...,1).

« Last Edit: Jun 21^{st}, 2008, 12:56am by Eigenray » 
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wonderful
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Re: n = (x^2+y^2+z^2)/(xyz)
« Reply #8 on: Jun 21^{st}, 2008, 5:34pm » 
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Thanks Eigenray so much for this great discussion. I just looked at your original solution and not quite follow the following: Quote:Let (x,y,z) be a solution with, WLOG, x <= y <= z, and z minimal. 
 if x<=y<=z then why z is minimal? Can you explain a bit more? Have A Great Day!


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Eigenray
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Re: n = (x^2+y^2+z^2)/(xyz)
« Reply #9 on: Jun 21^{st}, 2008, 5:59pm » 
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on Jun 21^{st}, 2008, 5:34pm, wonderful wrote:if x<=y<=z then why z is minimal? 
 I just mean pick a solution with max(x,y,z) as small as possible. It would probably be better to say x+y+z as small as possible, so we don't have to worry about whether the values are distinct. Call a solution minimal if it can't be transformed into one with a smaller value of x+y+z by changing a single value. Then through a series of steps, any solution can be transformed into a minimal one, but a priori a minimal solution need not be unique. The idea is that since xyz is cubic, it will be larger than x^{2}+y^{2}+z^{2} unless one number is much larger than the others. But if that's the case, we can always replace the largest number with a smaller one (until x,y,z form an acute triangle).

« Last Edit: Jun 21^{st}, 2008, 6:00pm by Eigenray » 
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wonderful
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Re: n = (x^2+y^2+z^2)/(xyz)
« Reply #10 on: Jun 21^{st}, 2008, 6:52pm » 
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Thanks Eigenray. This is very nice indeed! Have A Great Day!


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