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Find all positive integers $n$ for which the equation

$$ x + y + u + v = n \sqrt{ xyuv } $$

has a solution in positive integers.

This problem is taken from Vietnamese Mathematical Olympiad, 2002, Question B5

Context: Will Jaggy requested this problem and solution to be posted. See his comment

Calvin Lin
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    thank you. My first impression is that this is more specific than what I expected. – Will Jagy Mar 28 '20 at 19:14
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    I’m voting to close this as needing more details. Personally I don’t find a plain random question, without any commentary, and a screenshot of an answer very intriguing or worthy of being posted, but others may disagree. – gen-ℤ ready to perish Mar 28 '20 at 19:19
  • @gen-zreadytoperish I put in a fair amount of time on a question that was self-deleted. At my request, Calvin put both items here so that I could view them. There are some surprising claims in the answer. Oh, http://zakuski.utsa.edu/~jagy/Hurwitz_A_1907.pdf gives very good detail on Vieta type methods. – Will Jagy Mar 28 '20 at 19:26
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    (I do not care if this is deleted or closed. Will can get to it and view it even if it's deleted, which was my intention. Hence the context given above + making it community wiki.) – Calvin Lin Mar 28 '20 at 19:26
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    Don’t delete it just because one person didn’t like it! This is a democracy, after all $\ddot\smile$. Notice I didn’t downvote. – gen-ℤ ready to perish Mar 28 '20 at 19:27
  • Please do not use titles consisting only of math expressions; these are discouraged for technical reasons -- see Guidelines for good use of $\LaTeX$ in question titles. – mrtaurho Mar 28 '20 at 19:28
  • @mrtaurho Noted. Read the link. (Sorry, I haven't posted such questions much here, so am unaware of what to do in such cases. Thanks for the edit.) – Calvin Lin Mar 28 '20 at 19:30
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    @WillJagy This solution also applies to the similar 3-variable version. I've added a solution, which follows this. – Calvin Lin Mar 28 '20 at 19:31
  • Can you please edit the question in order to explain your context? Comments are fundamentally ephemeral, and can be deleted without notification (if, say, the comment is flagged for deleting for some reason, or if the post to which a comment is attached is deleted). – Xander Henderson Mar 29 '20 at 00:33

2 Answers2

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Solution from Selected Problems of the Vietnamese Mathematical Olympiad (1962–2009). (Volume 5, Mathematical Olympiad Series)

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Calvin Lin
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here is my list of (x+y+z+t)^2 = kxyzt in positive integers. Note k need not be a 

square.


ordered and fundamental integers x >= y >= z >= t >= 1,

( x+y+z+t)^2 = kxyzt   and 2(x+y+z+t)  <= kyzt:

      k               x      y      z      t
      1              10     10      9      1  2sum: 60 kyzt:  90
      1              12      6      4      2  2sum: 48 kyzt:  48
      1              15     10      3      2  2sum: 60 kyzt:  60
      1              18      9      8      1  2sum: 72 kyzt:  72
      1              21     14      6      1  2sum: 84 kyzt:  84
      1              30     24      5      1  2sum: 120 kyzt:  120
      1               4      4      4      4  2sum: 32 kyzt:  64
      1               6      6      3      3  2sum: 36 kyzt:  54
      1               8      5      5      2  2sum: 40 kyzt:  50

      2              10      5      4      1  2sum: 40 kyzt:  40
      2              12      8      3      1  2sum: 48 kyzt:  48
      2               4      3      3      2  2sum: 24 kyzt:  36
      2               8      4      2      2  2sum: 32 kyzt:  32

      3               4      4      3      1  2sum: 24 kyzt:  36
      3               6      2      2      2  2sum: 24 kyzt:  24
      3               9      6      2      1  2sum: 36 kyzt:  36

      4               2      2      2      2  2sum: 16 kyzt:  32
      4               6      3      2      1  2sum: 24 kyzt:  24

      5              10      8      1      1  2sum: 40 kyzt:  40
      5               5      2      2      1  2sum: 20 kyzt:  20

      6               6      4      1      1  2sum: 24 kyzt:  24

      8               4      2      1      1  2sum: 16 kyzt:  16

      9               2      2      1      1  2sum: 12 kyzt:  18

     12               3      1      1      1  2sum: 12 kyzt:  12

     16               1      1      1      1  2sum: 8 kyzt:  16
Will Jagy
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    I suspect they used $ k = n^2$ to make it easier to find the solutions once there's a bound on $n$. – Calvin Lin Mar 28 '20 at 19:34
  • @CalvinLin I think I have found the difference. Hurwitz minimizes the sum of variables subject to being in the same tree. Your Vietnam authors are taking the smallest sum out of all trees with the same multiplier, so they get just one. Yes, at the end they print the roots of the specific trees chosen. That suffices for finding the possible coefficients. Alright, I'm getting there. Thanks for the post. – Will Jagy Mar 28 '20 at 19:43
  • and, had they allowed coefficient 3, they would lose uniqueness for the smallest sum, as sum 12 (double sum 24) has either root 4431 which is primitive, or 6222 of which all are even – Will Jagy Mar 28 '20 at 19:49
  • (At the olympiads) It's pretty common to focus on 1 tree and then see how it grows. What's unique in their usage of Vieta, is focusing on the smaller solution in order to get the bounding on $n$. Another unique-ish application of Vieta's is in this solution (deleted to hide it), which demonstrates that the solution is unique. – Calvin Lin Mar 28 '20 at 20:11
  • @CalvinLin very nice. I noticed that problem but did not relate it to $x^3 + y^3 + z^3 - 3xyz$ – Will Jagy Mar 28 '20 at 20:16
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    Actually, that relation wasn't necessary. The conditions give $ p \mid q^3 + 1 = (q+1)(q^2 - q + 1)$ (and vice versa). The difficult case is dealing with $ p \mid q^2 - q + 1 , q \mid p^2 - p + 1$. – Calvin Lin Mar 28 '20 at 20:19
  • @Calvin, in.case of interest: for the Vietnam problem in $m$ variables, it already seems that the root (Hurwitz fundamental solution) with largest $x_1$ has multiplier $1,$ then $x_1 = 5m+10, ; ;$ $x_2 = 4m + 8, ; ;$ $x_3 = 5, ; ;$ $x_4 = \cdots= x_m = 1.$ Both sides of the equation are $100 (m+2)^2$ – Will Jagy Mar 30 '20 at 00:08