Number as sum of two squares

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savinov's blog

Number as sum of two squares

By savinov, 13 years ago, translation, In English

Hi, everybody! Recently, I tried to solve a problem. In this problem I had to determine, is number N sum of two squares? i.e. N = a² + b²?, where a and b are integer. N < 10¹⁵. Firstly, I tried to brute-force a and check N - a² for a square, but got TL. Also I tried to factorize the number and check degree of prime divisors p = 4k + 3 and used Fermat-Euler theorem to check the fact. But also I got TL. I suppose this problem to solve with number of operations $\text{[math]}$ . And I'm interested in technical realization of it. Can anybody show how to write accurate brute-force or factorization for example on C/C++? And what advices can you give to write same programs, making as less operations as possible? It's the problem from acm.timus.ru link here Thanks in advance!

number theory, realization

savinov
13 years ago
23

Comments (22)

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wil93

13 years ago, # |

How can you use an algorithm that checks N = a²+b² in the problem you linked? It asks for the minimum number of squares that sum to N, am I right?

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ibra

13 years ago, # ^ |

a little hint

http://en.wikipedia.org/wiki/Lagrange%27s_four-square_theorem

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Tranvick

13 years ago, # ^ |

← Rev. 3 →

+10

Any natural number can be represented as the sum of four integer squares. We check that N = a². If not, we check that N = a² + b². If this is not correct, we check that N = a² + b² + c². And the answer is 4, if this is also not correct.

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Kostroma

13 years ago, # ^ |

+14

how can we check if N can be represented as a sum of three squares fast?

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niklasb

13 years ago, # ^ |

← Rev. 2 →

The following question on MathOverflow might help: http://mathoverflow.net/questions/57981/testing-whether-an-integer-is-the-sum-of-two-squares

EDIT: Sorry, I misread your comment :)

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OSt

13 years ago, # ^ |

← Rev. 2 →

-15

IMHO, in that problem more easy to check if N can be represented as sum of 1,2,4 squares. Else — 3. :-)

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afix

13 years ago, # ^ |

← Rev. 2 →

27 = 1*1 + 1*1 + 3*3 + 4*4 = 1*1 + 1*1 + 5*5
27 != a*a
27 != a*a + b*b
Fail.

→ Reply

OSt

13 years ago, # ^ |

if N != (8*t+7)*4^k then there are 3 squares.

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Kostroma

13 years ago, # ^ |

← Rev. 3 →

oh, don't you know the proof or some link with the proof?

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riadwaw

13 years ago, # ^ |

+13

And now read your previos explanation.

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Kostroma

13 years ago, # ^ |

← Rev. 3 →

-8

sorry, I've misunderstood you

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Tranvick

13 years ago, # ^ |

int three_square(ll x){
	while (!(x&3)) x>>=2;
	return (x+1)%8;
}

It returns 0, if x is not a sum of three squares.

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CherryTree

13 years ago, # ^ |

How come?

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Tranvick

13 years ago, # ^ |

This is a very difficult method of pen and paper :)

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wil93

13 years ago, # ^ |

I think that algorithm follows this: http://www.proofwiki.org/wiki/Integer_as_Sum_of_Three_Squares

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jaineshp

9 years ago, # ^ |

-12

For sum of two squares, how can it be modified?

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ha1vanka

9 years ago, # ^ |

+29

Don't answer him. This is one of the problems from a running contest.

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jaineshp

9 years ago, # ^ |

-14

Sorry?

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I_will_win_a_div1_round

9 years ago, # ^ |

-27

Motori

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red_coder

13 years ago, # |

← Rev. 2 →

according to me we can run a loop in which 'i' goes from 1 to sqrt(N/2) and if N-i*i is a perfect square we are done....

for(int i=1; i*i<=N/2; i++) { if((N-i*i) is a perfect square) done; }

its complexity is O(sqrt(N)).....

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savinov

13 years ago, # ^ |

How do you check number for square? I did the same and used sqrt() and got TL.

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Oi.

13 years ago, # |

An arbitrary positive number n is expressible as the sum of two squares iff, given its prime factorization

n=p1^(a1)p2^(a2)p3^(a3)...pk^(ak),

none of pi^(ai) +1 is divisible by 4 (Conway and Guy 1996, p. 147). http://mathworld.wolfram.com/SquareNumber.html

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