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 [tex]a[/tex] the Diophantine equation [tex]\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{a}{xyz}[/tex] has infinitely many integer solutions [tex](x,y,z)[/tex] such that [tex]\gcd(a,xyz)=1[/tex]. I have not solved it, I have just found it interesting and suspect that others on this forum might find it interesting as well. 
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] 
[spoiler]
ZetaFlux 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] 
nm

William,
Nice solution. You might consider submitting it to the AMM for publication. 
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=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 nonsubscribers.  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 
[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 nonsubscribers.
William[/QUOTE] I'll PM you with the information. 
[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 nonsubscribers.
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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