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How to find instances where d(a,b)=p2 for p a prime

asked 11 years ago

JoshIzzard gravatar image

Suppose I have a dimension formula (for a Lie algebra representation) given by dima,b=(a+1)(b+1)(a+b+2)2. I now would like to find pairs (a,b) where dima,b=p2 for p a prime? What are some techniques for accomplishing this? Should I first filter out a list of primes using isprime and then check possible pairs (a,b) for each prime p<N, say 1000.

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Are you sure there are any such pairs to find? Unless I'm missing something, you can't fit 2p2 into (a+1)(b+1)(a+b+2) successfully.

DSM gravatar imageDSM ( 11 years ago )

Ah @DSM this may indeed be the case. Now I suppose I would like to find points where d(a,b)=n for nN. Does Sage have a toolkit that allows me to extract integral points on the surface F(a,b,n)=d(a,b)n=0?

JoshIzzard gravatar imageJoshIzzard ( 11 years ago )

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answered 11 years ago

tmonteil gravatar image

updated 11 years ago

I am not sure i understand your question correctly. If you want to iterate over the pairs of integers, you can have a look at multidimensional enumeration:

sage: f = lambda a,b : (a+1)*(b+1)*(a+b+2)
sage: n = 720

sage: for a,b in xmrange([n,n]):
sage:     if f(a,b) == 2*n:
sage:         print a,b      
7 9
9 7

And actually you can be more restrictive and ask a and b to be less than floor(sqrt(2*n)) (instead of n):

sage: bound = floor(sqrt(2*n))
sage: for a,b in xmrange([bound,bound]):
sage:     if f(a,b) == 2*n:
sage:         print a,b      
7 9
9 7

Hence, you get all answers in time O(n) instead of O(n2). But why not solving directly your equation ? Unfortunately, it seems that the solver is not able to solve the diophantine equation f(a,b) == 2*n directly:

sage: var('a','b')
(a, b)
sage: solve([f(a,b) == 2*n ], a, b)
([a == -1/2*(4*b + sqrt(b^4 + 4*b^3 + 6*b^2 + 5764*b + 5761) + b^2 + 3)/(b + 1), a == -1/2*(4*b - sqrt(b^4 + 4*b^3 + 6*b^2 + 5764*b + 5761) + b^2 + 3)/(b + 1)],
 [1, 1])

But you can still use Sage solver when a is fixed:

sage: var('b')
b
sage: assume(b, 'integer')
sage: assume(b >= 0)

sage: for a in xrange(bound):
sage:     sol = solve([f(a,b) == 2*n],b)
sage:     if len(sol) != 0:
sage:         print a, sol[0].rhs()
sage:         
7 9
9 7

Now, you get an answer in time O(n), which is not completely satisfactory (one could expect a polynomial in log(n)), but not that bad.

That said, since the solver is quite slow, the multiplicative constant in the complexity is quite huge and the previous method is faster for small n (this is clear for n=6216 : 42.7ms vs 1.11s). The last method becomes faster only later (this is clear for n=3065451 : 94.43s vs 29.77s).

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@tmontneil Thank you for taking the time to write out this detailed solution. You have given me an idea as to how to move forward with this for more detailed dimension formulae. Eventually I hope to write an algorithm that checks for dimension equal to p2 for the type D2k+1 Lie algebras as well, but that dimension formula is quite a bit more complicated (if you are interested in the subject see Fulton and Harris p. 410 exercise 24.42). Thanks again for your help. Regards

JoshIzzard gravatar imageJoshIzzard ( 11 years ago )
1

answered 11 years ago

vdelecroix gravatar image

updated 11 years ago

Just to improve tmonteil answer you can use simple nested loops to enumerate solutions with the same "complexity" using the fact that your function d(a,b) is increasing in both variables:

N = 2*n
a = floor(sqrt(N-1))
b = 0
while a != -1:
    while (a+1)*(b+1)*(a+b+2) < N:
        b += 1
    if (a+1)*(b+1)*(a+b+2) == N:
        print a,b
    else:
        b -= 1
    a -= 1

and then, computing the solution for N=2*3065451 runs in 8.33 ms which looks like a thousand times better !

And I am pretty sure that it is possible to do something more clever than increasing/decreasing by units in my function above...

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@vdelecroix thanks very much, this is quite impressive

JoshIzzard gravatar imageJoshIzzard ( 11 years ago )

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Asked: 11 years ago

Seen: 565 times

Last updated: May 25 '13