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let $a,b,c,x,y,z$ be all pairwise coprime integers . Show that: $$abx^2+bcy^2+acz^2=(xyz)^2+2abc$$ has no integral solutions if $a,b,c,x,y,z >1$. I tried to confirm the results in wolfram but I am totally clueless as to how to prove this. Any hints?

Tito Piezas III
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    I think you posted this recently; I don't know what you mean by nontrivial; mostly, i don't know where you got the problem. Note: $x^2 + y^2 + z^2 = 3xyz$ has infinitely many solutions in positive integers. The related $x^2 + y^2 + z^2 = Axyz + B$ will have infinitely many solutions if any. Your term $(xyz)^2$ is so large that you cannot have two out of three variables large and one small, so Markov's setup does not apply. For any triple $(a,b,c)$ I imagine the set of solutions is finite. But again, where did you get this? – Will Jagy Dec 20 '15 at 00:57
  • I edited the question for clarification. I vaguely remember seeing this problem in an Algebra book from Singapore regarding symmetric Diophantine equations. I checked out on wolfram the possible solutions and noted indeed there are no solutions. I am puzzled as to why it is so. I was expecting infinitely many solutions. But you pointed out something very interesting regarding the size of $(xyz)^2$.Perhaps, that's the reason. I'll look at the problem from this angle. –  Dec 20 '15 at 11:28
  • @Ramunjndscpl: I changed a phrase since you could not have confirmed your conjecture as it is incorrect. See W. Jagy's counter-example. – Tito Piezas III Dec 21 '15 at 04:44

3 Answers3

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your problem is wrong

  a   b   c      x   y  z
 49  39  25     41  11  4  


Sun Dec 20 13:08:27 PST 2015
Tito Piezas III
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Will Jagy
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  • I guess Wolfram needs an upgrade. it's misleading. Thanks Will Jagy. –  Dec 21 '15 at 17:09
  • @Ramunjndscpl actually you need to get an ordinary procedural language and learn how to use it. People seem to like Python, and then there is Sage. I use C++. You should not be posting false statements and getting people to waste time on them as thought they were true. – Will Jagy Dec 21 '15 at 17:53
  • @WillJagy, it is not a false statement. Please note in the stem, it's stated as a question. The reason why I thought there were no actual solutions it's because I was mislead by Wolfram. –  Dec 21 '15 at 23:59
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    @Ramunjndscpl: This type of multi-variable Diophantine equation is beyond Walpha. It can answer, for example, "solve x^2-5y^2=1 over the integers". But not "solve a^2+b^2+c^2 = d^2 over the integers". In fairness to Walpha, solving arbitrary multiple-variable Diophantine equations is an intrinsically hard problem, and it is not something one can expect to find free over the Internet. – Tito Piezas III Dec 23 '15 at 13:15
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    @WillJagy: Individ found a nice parameterization that, for carefully chosen parameter, yield prime $a,b,c,x,y,z$. A quick search already yielded 220 solutions, such as, $$293,, 17389,, 61,, 11,, 2,, 7\37493,, 2286589,, 61,, 11,, 2,, 193$$ and so on. I wouldn't be suprised if the system turned out to have an infinite number of pairwise co-prime solutions. – Tito Piezas III Dec 23 '15 at 15:35
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Note that for example $(a,b,c,x,y,z) = (1,1,7,1,5,3)$ is a solution to this equation.

So, what do you mean by "non-trivial"?

user133281
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    Computer search suggests that there are no solutions with all numbers different from $1,0,-1$. I suspect this is what was meant. – Wojowu Dec 20 '15 at 10:51
  • @user133281, I edited the question for clarification. I am however pleasantly surprised at your answer. I wonder why Wolfram did not provide such answer. –  Dec 20 '15 at 11:35
  • @Wojowu $(49, 39, 25, 41, 11, 4)$ – Will Jagy Dec 20 '15 at 21:14
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    @WillJagy It seems that I was just close of finding a solution - I went with $a<20,b<40,c<60$. – Wojowu Dec 20 '15 at 21:48
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    @Wojowu I started by allowing $\gcd(c,x) > 1$ and found some worthwhile sextuples for this revised problem with somewhat smaller entries. It seemed worth it to raise the bounds, although I was not really expecting a solution to the problem with all gcd's required to be $1.$ It's a habit from decades of this, I weaken restrictions to get a little output from the computer. Frustrating, and uninformative, seeing nothing at all. – Will Jagy Dec 20 '15 at 21:59
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$$abx^2+bcy^2+acz^2=(xyz)^2+2bca$$

$$x=2n+1$$

$$c=2n^2+2n+1$$

$$a=(2n^2+2n+1)y^2+z^2$$

$$b=4n^2(n+1)^2y^2+(2n^2+2n+1)z^2$$

individ
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  • @invidid, you always have the weirdest parametric solutions. Thanks for your input. –  Dec 21 '15 at 17:07
  • @individ: He wants it pairwise coprime. That means, if you have a set of $n$ integers $x_1, x_2, x_3,\dots x_n$, then all possible pairs $x_i, x_k$ must not share any factor. – Tito Piezas III Dec 23 '15 at 13:24
  • @TitoPiezasIII How can there be problems? You just have to choose the numbers are what we need. For example. $n=2$ ; $x=5$ ; $y=2$ ; $z=7$ ; $c=13$ ; $a=101$ ; $b=1213$ . – individ Dec 23 '15 at 13:51
  • @individ: That's very good, all $a,b,c,x,y,z$ are primes, hence are automatically pairwise co-prime. You should edit your answer to include that example, and maybe a few others also. – Tito Piezas III Dec 23 '15 at 14:09