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 philmoore 2008-04-30 22:52

Diophantine problem

This problem was posed in the April 2008 issue of [I]The American Mathematical Monthly[/I] by Jeffrey Lagarias, U. of Michigan:

Determine for which integers $$a$$ the Diophantine equation
$$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{a}{xyz}$$
has infinitely many integer solutions $$(x,y,z)$$ such that $$\gcd(a,xyz)=1$$.

I have not solved it, I have just found it interesting and suspect that others on this forum might find it interesting as well.

 Zeta-Flux 2008-05-01 13:51

Here is a partial solution.

[SPOILER]It is easy to reduce the equation to xy+xz+yz=a.[/SPOILER]

[SPOILER]It is also easy to see that 2 cannot divide a.[/SPOILER]

[SPOILER]Let x=-(4k^2+2k+a), y=8k^2+2k+2a, z=8k^2+6k+2a+1.[/SPOILER]

[SPOILER]We compute that xy+xz+yz=a for any k.[/SPOILER]

[SPOILER]On the other hand, taking k==-1 mod a, we see that x==-2 mod a, y==6 mod a, and z==3 mod a.[/SPOILER]

[QUOTE]Note that as k increases (for large enough k), x,y,z also increase in magnitude.[/QUOTE]

[SPOILER]Thus, this solution works for any a not divisible by 2 or 3.[/SPOILER]

 wblipp 2008-05-03 23:20

[spoiler]
Zeta-Flux has already observed that a cannot be even (he said it was easy to see, but it took me a while to see that it's because x and y and z must all be odd by the gcd requirement, and xy + xz + yz would then be odd)

let a = 2b+1

x = -(2b+1)^2 * c^2 * (18b+10) - 3*(2b+1)*c*(b+1) + b
y = (2b+1)^2 * c^2 * (54b+30) - (2b+1)*c*(9*b+1) + 1
z = (2b+1)^2 * c^2 * (27b+15) + (2b+1)*c*(9*b+7) + 1

Gives an infinite family of solutions, and the gcd requirement is met because

x = b mod (2b+1)
y = 1 mod (2b+1)
z = 1 mod (2b+1)

Thus there is an infinite family of solutions for all odd a.

[/spoiler]

 ATH 2008-05-04 02:22

nm

 Zeta-Flux 2008-05-05 01:51

William,

Nice solution. You might consider submitting it to the AMM for publication.

 philmoore 2008-05-05 03:31

I second Pace's suggestion, but I am still trying to understand the process of how both of you come up with these beautiful parametrized solutions. I see this sort of thing occasionally on the NMBRTHRY listserve, but I have no idea how people come up with them! Care to share any secrets?

 wblipp 2008-05-05 04:10

[QUOTE=philmoore;132752]I second Pace's suggestion, but I am still trying to understand the process of how both of you come up with these beautiful parametrized solutions. I see this sort of thing occasionally on the NMBRTHRY listserve, but I have no idea how people come up with them! Care to share any secrets?[/QUOTE]

Wouldn't they want submitters to be subscribers - or at least regular readers? I don't know how to submit it without at least reading - the web doesn't seem to offer that information to non-subscribers.

-----------------

I started with x=b, y=z=1, and hoped I could extend it. A quadratic extension would give me six "free" variables and only four constraints, so I started looking for solutions to the multiplication for

x = x1*s^2 + x2*s + b
y = y1*s^2 + y2*s + 1
z = z1*s^2 + z2*s + 1

The trickiest part was finding integer solutions for the s^4 term,
x1*y1+x1*z1+y1*z1=0

x1 = - y1 * z1 / (y1 + z1)

I did a quick search on small values of y1 and z1 looking for integer values of x1. I ran into a dead end with y1 = z1 = 2, so I tried the smallest "off diagonal" solution, y1=3 z1=6. I think Pace used this same case. I imagine other families of solutions can be made from the other off diagonal solutions.

I set x1=-2d, y1=3d, z1=6d and then looked for similar tricks to force the s^3, s^2, and s terms of the product to zero. I used the symbolic manipulation package in Mathcad (a Maple variant) at each step to express the product as a polynomial in s, then looked for a way to force one more coefficient to be zero - substitute that into the definition and repeat.

Once I had a parameterized solution to the multiplication, I made "s" a multiple of (2b+1) to easily enforce the gcd condition.

William

 philmoore 2008-05-05 21:30

[QUOTE=wblipp;132755]Wouldn't they want submitters to be subscribers - or at least regular readers? I don't know how to submit it without at least reading - the web doesn't seem to offer that information to non-subscribers.
William[/QUOTE]

I'll PM you with the information.

 m_f_h 2008-05-19 14:17

[quote=wblipp;132755]Wouldn't they want submitters to be subscribers - or at least regular readers? I don't know how to submit it without at least reading - the web doesn't seem to offer that information to non-subscribers.
William[/quote]

Indeed this is a pity. And even having an institutional subscription to JSTOR I'm often frustrated on that because the "moving wall" is quite far away in the past :-( !

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