20090110, 23:32  #1 
Dec 2008
you know...around...
2^{3}·101 Posts 
Short question
Now, we all know that if n=4m+1 is prime it can be expressed as the sum of two squares in only one way.
I was wondering if this is also true the other way round, i.e. if n=4m+1 can be written as the sum of two squares in only one way (Edit: and gcd(a;b)=1), does that mean n is prime? And if so, how efficient would that test be? I mean, checking the possibilities of a²+b²=n. Surely not as fast as APRCL or ECPP, but I think it could be more efficient than trial division? Last fiddled with by mart_r on 20090110 at 23:41 
20090111, 00:44  #2  
Aug 2006
1763_{16} Posts 
Quote:
Checking each possibility as the sum of two squares takes O(sqrt (n) * log n). 

20090111, 01:04  #3  
Dec 2008
179 Posts 
Quote:


20090111, 16:06  #4  
Dec 2008
you know...around...
2^{3}×101 Posts 
Aha. Good to know.
Quote:
And surely there are even finer techniques to find sums of squares. But I don't know very much about it. Last fiddled with by mart_r on 20090111 at 16:08 

20090111, 17:57  #5 
Aug 2006
5987_{10} Posts 
If you only need to test 1% of the possibilities, that's O(sqrt n * log n).

20090111, 19:00  #6 
Dec 2008
you know...around...
2^{3}·101 Posts 

20090111, 19:12  #7 
Aug 2006
5987_{10} Posts 
Basically, yes. The dominant term is sqrt(n), just like for trial division. Both have a large number of possible optimizations, so it's not clear which would be practically faster, but neither could be used for large numbers.

20090111, 23:00  #8 
"Phil"
Sep 2002
Tracktown, U.S.A.
3·373 Posts 
Euler, for the record, did use this as a primality proving method. In practice, you can sieve out which nx^2 are not squares modulo 2, 3, 5, 7, etc. very quickly. In the end, you need test very few numbers of this form to see if they are actually perfect squares. I do not have the details at hand, but I recall studying this problem a few years ago and coming to the conclusion that finding all representations of n as a sum of squares x^2+y^2 was of approximate order O(n^1/4), although an extra factor of ln n or so would not surprise me. Obviously, we could use this as either a primality proving method or a factoring method.

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