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