20080430, 22:52  #1 
"Phil"
Sep 2002
Tracktown, U.S.A.
3·373 Posts 
Diophantine problem
This problem was posed in the April 2008 issue of The American Mathematical Monthly by Jeffrey Lagarias, U. of Michigan:
Determine for which integers the Diophantine equation has infinitely many integer solutions such that . I have not solved it, I have just found it interesting and suspect that others on this forum might find it interesting as well. 
20080501, 13:51  #2  
May 2003
11000001011_{2} Posts 
Here is a partial solution.
It is easy to reduce the equation to xy+xz+yz=a. It is also easy to see that 2 cannot divide a. Let x=(4k^2+2k+a), y=8k^2+2k+2a, z=8k^2+6k+2a+1. We compute that xy+xz+yz=a for any k. 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. Quote:


20080503, 23:20  #3 
"William"
May 2003
New Haven
3·787 Posts 
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. 
20080504, 02:22  #4 
Einyen
Dec 2003
Denmark
2×29×53 Posts 
nm
Last fiddled with by ATH on 20080504 at 02:26 
20080505, 01:51  #5 
May 2003
7·13·17 Posts 
William,
Nice solution. You might consider submitting it to the AMM for publication. 
20080505, 03:31  #6 
"Phil"
Sep 2002
Tracktown, U.S.A.
3·373 Posts 
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?

20080505, 04:10  #7  
"William"
May 2003
New Haven
3×787 Posts 
Quote:
 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 Last fiddled with by wblipp on 20080505 at 04:21 Reason: fix a sign 

20080505, 21:30  #8 
"Phil"
Sep 2002
Tracktown, U.S.A.
3·373 Posts 

20080519, 14:17  #9 
Feb 2007
2^{4}×3^{3} Posts 
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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