Having some trouble finding representations for given norms with the integers in the complex quadratic fields in 1 (mod 4)…

 

This started to show itself in  …

 

 

 

 

 

The problem is now evident…  That is, the equation  has no solution in integers, yet the equation  does have solutions in integers…  I believe the problem may be that  covers all the points in the maximal integral domain of , whereas  doesn’t include the exceptional points, which in some cases, fixes unique factorization.  Why this didn’t show up before I started working in  … I’m not sure yet.

 

I believe I’ve found a couple solutions to this problem… I’ll present the solution I feel is most suitable for hand calculation first, which is also the one I personally like best.

 

First of all, one needs to see the relationship between  and  

 

 

Since it’s more natural, and probably easier, to solve eq. (1), let’s find an equivalent equation in u, v for eq. (1), solve this equation, and then use relations (3) to map that solution back to eq. (2).

 

So, in general, we want to find solutions to this equation… .  Once having found the solutions to this equation, we can find the solutions to (2) by rearranging relations (3):

 

 

 

(I’ve yet to prove it, but it appears the +v corresponds to the +v for both a and b… and vice-versa for v.  This probably follows because +v and v are inverses mod 4n, and then, u+v and u-v are inverses mod 2n… I think…?)

 

So, returning to the problem above:

 

 

 

However, now let’s solve the related equation .  So, now:

 

 

 

 

 

Yes… one could use Tonelli-Shanks.  However, I’ve found that this method is usually easier with moderately sized numbers… well, numbers of the size I work with, anyway…

 

The other method I’ve found treats  as a general quadratic Diophantine equation in two variables.  I derived this method from Dario Alpern’s methods page for his Generic Two integer variable equation solver (If you’ve never visited Dario’s site before, you’re in for quite a treat…  However, do be cautioned that you know what they say about there always being someone who’s better than you… no matter what subject it is… well, that’s Dario with stuff like this…)

 

It appears that  will always be an elliptical case of the general quadratic Diophantine equation in two variables.  This follows easily from investigation of the possible discriminants.

 

So, modifying the equation on Dario’s method page to suit this application, I came up with:

 

 

and…

 

 

 

are the corresponding values of b which must be checked to see if they’re integral.

 

It appears that it’s probably only necessary to check all possible values of b for  since there appears to be a rather obvious correspondence between the negative and positive values… although I’ve yet to prove this.

 

If we apply formula (4) in the case of our problem above…

 

 

 

we get:

 

 

 

indicating we need to check the corresponding values of b for  with equation (5).

 

This is much easier done with an Excel worksheet… http://paul-mccarthy.us/Misc/Book1.htm

 

The first row below the a values in the Excel worksheet correspond to taking the plus sign in front of the radical in (5), and the second row below the a values corresponds to taking the negative sign in front of the radical.

 

As can be seen, this gives essentially the same results as the first method, including all the plus-minus combinations (if one includes the negative a values.

 

 

The second method does have certain advantages,… however, the numbers generally don’t seem so pleasant.  One needs to calculate square roots… although only the integral portion… and the numbers get much larger.  If I knew of a quick and easy way to determine whether a particular number was a perfect square…  However, other than for machine implementation using congruences to various moduli, I do not know of any methods suitable for hand computation.  These are some of the disadvantages with the second method in regards to hand computation.

 

However, I’m working on attempting to combine the two methods.  Perhaps some of the possible solutions that the second method comes up with can be thrown out by using some congruence methods similar to the first method… probably using elements of the first method to ease computation in the second method.

 

From what I know at this time, I believe the first method is probably better for hand computation.