question on execution time for qfbsolve

[American Citizen <website.reader3@gmail.com>]

Hello:

Currently I am using qfbsolve(Qfb(1,0,1),N,3) to find pairs of squares 
summing to N (if possible)

The web page 
https://pari.math.u-bordeaux.fr/dochtml/html/Arithmetic_functions.html#qfbsolve 
states that "The integer n can also be given by its factorization matrix 
fa = factor(n) or by the pair [n, fa]."

Currently my values for N < 1,000,000 and in fact are only primes or 
composites of primes of 1 mod 4 only.

How much of a speed up would I obtain by doing 
qfbsolve(Qfb(1,0,1),[N,factor(N)],3)  instead?

Randall