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Aug 12 at 19:27 comment added gnasher729 I looked at the problem “are there a, b >= 0 such that a^5+b^5=s” and inspired by this thread, it turns out that a+b divides a^5+b^5. Which may help finding solutions.
Oct 23, 2014 at 18:28 comment added Max Alekseyev Well, at least for Blum integers mentioned above, there are no square factors, and no cancelling is required.
Oct 23, 2014 at 15:15 comment added ndkrempel But "cancel all possible square factors" is not known to be polynomial time, so you have not demonstrated the equivalence.
Oct 21, 2014 at 16:48 comment added Max Alekseyev @ndkrempel: Strictly speaking, equivalence here is with being the sum of two coprime squares. For non-coprime squares, we can try to cancel all possible square factors of $n$ and then test if it becomes the sum of coprime squares. In particular, for $n=45$, we have that -1 is a square modulo $45/3^2=5$.
Oct 20, 2014 at 21:22 comment added ndkrempel Can you explain your equivalence? It is not directly equivalent as, for example, 45 is a sum of two squares but -1 is not a square modulo 45.
Jul 19, 2011 at 15:45 comment added Max Alekseyev Why "quadratic residuosity problem is supposed to be hard when n is a Blum integer"? It is formulated for a generic composite number. A Blum integer would be a particular case of the problem.
Mar 10, 2011 at 11:02 comment added Emil Jeřábek To expand on Peter’s comment, the quadratic residuosity problem is supposed to be hard when $n$ is a Blum integer, in which case we know a priori that $-1$ is not a square.
Mar 9, 2011 at 23:32 comment added Peter Shor But is testing whether -1 is a square as hard as testing whether an arbitrary residue (with Jacobi symbol 1) is a square? Testing whether an arbitrary residue is a square is the supposedly hard quadratic residuosity problem.
Mar 9, 2011 at 19:51 history answered Max Alekseyev CC BY-SA 2.5