Lagrange four-squares theorem — deterministic complexityLagrange four-squares theorem: efficient algorithm with units modulo a prime?The Green-Tao theorem and positive binary quadratic formsEuler and the Four-Squares TheoremApplications of finite continued fractionsWhat are the consequences of a polynomial time algorithm for finding out if a given number is expressible as the sum of two squares?A 'generalized Four Squares Theorem'?Polynomial-time complexity and a question and a remark of SerreFinding integer representation as difference of two triangular numbersSeeking references for finding primes infinitely oftenLagrange four squares theorem
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Lagrange four-squares theorem — deterministic complexity
Lagrange four-squares theorem: efficient algorithm with units modulo a prime?The Green-Tao theorem and positive binary quadratic formsEuler and the Four-Squares TheoremApplications of finite continued fractionsWhat are the consequences of a polynomial time algorithm for finding out if a given number is expressible as the sum of two squares?A 'generalized Four Squares Theorem'?Polynomial-time complexity and a question and a remark of SerreFinding integer representation as difference of two triangular numbersSeeking references for finding primes infinitely oftenLagrange four squares theorem
$begingroup$
Lagrange's four-squares theorem states that every natural number can be represented as the sum of four integer squares. Rabin and Shallit gave a randomised algorithm that finds one of these solutions in quadratic time. My question is if anything is known about the deterministic time complexity of finding one of the solutions? Any pointers would be appreciated.
(It seems that enumerating all the solutions is hard as factoring in certain cases (via Jacobi's four-square theorem), but correct me if I am wrong.)
nt.number-theory computational-complexity sums-of-squares
edited Apr 20 at 9:45
Martin Sleziak
3,13032231
asked Apr 19 at 13:22
Occams_TrimmerOccams_Trimmer
1835
$endgroup$
add a comment |
$begingroup$
Lagrange's four-squares theorem states that every natural number can be represented as the sum of four integer squares. Rabin and Shallit gave a randomised algorithm that finds one of these solutions in quadratic time. My question is if anything is known about the deterministic time complexity of finding one of the solutions? Any pointers would be appreciated.
(It seems that enumerating all the solutions is hard as factoring in certain cases (via Jacobi's four-square theorem), but correct me if I am wrong.)
nt.number-theory computational-complexity sums-of-squares
edited Apr 20 at 9:45
Martin Sleziak
3,13032231
asked Apr 19 at 13:22
Occams_TrimmerOccams_Trimmer
1835
$endgroup$
1
$begingroup$
Your parenthetical remark is essentially correct. For instance, this is at least as difficult as factoring semiprimes, because it lets us compute $1+p+q+pq$ given a semiprime $pq$, and from $p+q,pq$ it's easy to compute $p,q$.
$endgroup$
– Wojowu
Apr 19 at 13:51
add a comment |
$begingroup$
Lagrange's four-squares theorem states that every natural number can be represented as the sum of four integer squares. Rabin and Shallit gave a randomised algorithm that finds one of these solutions in quadratic time. My question is if anything is known about the deterministic time complexity of finding one of the solutions? Any pointers would be appreciated.
(It seems that enumerating all the solutions is hard as factoring in certain cases (via Jacobi's four-square theorem), but correct me if I am wrong.)
nt.number-theory computational-complexity sums-of-squares
edited Apr 20 at 9:45
Martin Sleziak
3,13032231
asked Apr 19 at 13:22
Occams_TrimmerOccams_Trimmer
1835
$endgroup$
Lagrange's four-squares theorem states that every natural number can be represented as the sum of four integer squares. Rabin and Shallit gave a randomised algorithm that finds one of these solutions in quadratic time. My question is if anything is known about the deterministic time complexity of finding one of the solutions? Any pointers would be appreciated.
(It seems that enumerating all the solutions is hard as factoring in certain cases (via Jacobi's four-square theorem), but correct me if I am wrong.)
nt.number-theory computational-complexity sums-of-squares
nt.number-theory computational-complexity sums-of-squares
edited Apr 20 at 9:45
Martin Sleziak
3,13032231
asked Apr 19 at 13:22
Occams_TrimmerOccams_Trimmer
1835
edited Apr 20 at 9:45
Martin Sleziak
3,13032231
asked Apr 19 at 13:22
Occams_TrimmerOccams_Trimmer
1835
edited Apr 20 at 9:45
Martin Sleziak
3,13032231
edited Apr 20 at 9:45
Martin Sleziak
3,13032231
edited Apr 20 at 9:45
Martin Sleziak
3,13032231
3,13032231
asked Apr 19 at 13:22
Occams_TrimmerOccams_Trimmer
1835
asked Apr 19 at 13:22
Occams_TrimmerOccams_Trimmer
1835
asked Apr 19 at 13:22
Occams_TrimmerOccams_Trimmer
1835
1835
1
$begingroup$
Your parenthetical remark is essentially correct. For instance, this is at least as difficult as factoring semiprimes, because it lets us compute $1+p+q+pq$ given a semiprime $pq$, and from $p+q,pq$ it's easy to compute $p,q$.
$endgroup$
– Wojowu
Apr 19 at 13:51
add a comment |
1
$begingroup$
Your parenthetical remark is essentially correct. For instance, this is at least as difficult as factoring semiprimes, because it lets us compute $1+p+q+pq$ given a semiprime $pq$, and from $p+q,pq$ it's easy to compute $p,q$.
$endgroup$
– Wojowu
Apr 19 at 13:51
1
1
$begingroup$
Your parenthetical remark is essentially correct. For instance, this is at least as difficult as factoring semiprimes, because it lets us compute $1+p+q+pq$ given a semiprime $pq$, and from $p+q,pq$ it's easy to compute $p,q$.
$endgroup$
– Wojowu
Apr 19 at 13:51
$begingroup$
Your parenthetical remark is essentially correct. For instance, this is at least as difficult as factoring semiprimes, because it lets us compute $1+p+q+pq$ given a semiprime $pq$, and from $p+q,pq$ it's easy to compute $p,q$.
$endgroup$
– Wojowu
Apr 19 at 13:51
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
As far as I know, this is still an open problem. This is discussed in Section $5$ of the paper Finding the four squares in Lagrange's theorem by Pollack and Treviño. They mention that there is a deterministic polynomial-time algorithm when $n$ is a prime via quaterion multiplication, due to Bumby. Assuming a conjecture of Heath-Brown, there is a deterministic polynomial-time algorithm that works for all $n$. Finally, they mention that a positive proportion of all numbers can be written as the sum of four squares in deterministic polynomial time. Under the Extended Riemann Hypothesis, almost all numbers can be written as the sum of four squares in deterministic polynomial time.
answered Apr 19 at 14:36
Tony HuynhTony Huynh
19.9k672131
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$begingroup$
As far as I know, this is still an open problem. This is discussed in Section $5$ of the paper Finding the four squares in Lagrange's theorem by Pollack and Treviño. They mention that there is a deterministic polynomial-time algorithm when $n$ is a prime via quaterion multiplication, due to Bumby. Assuming a conjecture of Heath-Brown, there is a deterministic polynomial-time algorithm that works for all $n$. Finally, they mention that a positive proportion of all numbers can be written as the sum of four squares in deterministic polynomial time. Under the Extended Riemann Hypothesis, almost all numbers can be written as the sum of four squares in deterministic polynomial time.
answered Apr 19 at 14:36
Tony HuynhTony Huynh
19.9k672131
$endgroup$
add a comment |
$begingroup$
As far as I know, this is still an open problem. This is discussed in Section $5$ of the paper Finding the four squares in Lagrange's theorem by Pollack and Treviño. They mention that there is a deterministic polynomial-time algorithm when $n$ is a prime via quaterion multiplication, due to Bumby. Assuming a conjecture of Heath-Brown, there is a deterministic polynomial-time algorithm that works for all $n$. Finally, they mention that a positive proportion of all numbers can be written as the sum of four squares in deterministic polynomial time. Under the Extended Riemann Hypothesis, almost all numbers can be written as the sum of four squares in deterministic polynomial time.
answered Apr 19 at 14:36
Tony HuynhTony Huynh
19.9k672131
$endgroup$
add a comment |
$begingroup$
As far as I know, this is still an open problem. This is discussed in Section $5$ of the paper Finding the four squares in Lagrange's theorem by Pollack and Treviño. They mention that there is a deterministic polynomial-time algorithm when $n$ is a prime via quaterion multiplication, due to Bumby. Assuming a conjecture of Heath-Brown, there is a deterministic polynomial-time algorithm that works for all $n$. Finally, they mention that a positive proportion of all numbers can be written as the sum of four squares in deterministic polynomial time. Under the Extended Riemann Hypothesis, almost all numbers can be written as the sum of four squares in deterministic polynomial time.
answered Apr 19 at 14:36
Tony HuynhTony Huynh
19.9k672131
$endgroup$
As far as I know, this is still an open problem. This is discussed in Section $5$ of the paper Finding the four squares in Lagrange's theorem by Pollack and Treviño. They mention that there is a deterministic polynomial-time algorithm when $n$ is a prime via quaterion multiplication, due to Bumby. Assuming a conjecture of Heath-Brown, there is a deterministic polynomial-time algorithm that works for all $n$. Finally, they mention that a positive proportion of all numbers can be written as the sum of four squares in deterministic polynomial time. Under the Extended Riemann Hypothesis, almost all numbers can be written as the sum of four squares in deterministic polynomial time.
answered Apr 19 at 14:36
Tony HuynhTony Huynh
19.9k672131
edited 2 days ago
answered Apr 19 at 14:36
Tony HuynhTony Huynh
19.9k672131
answered Apr 19 at 14:36
Tony HuynhTony Huynh
19.9k672131
answered Apr 19 at 14:36
Tony HuynhTony Huynh
19.9k672131
19.9k672131
add a comment |
add a comment |
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1
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Your parenthetical remark is essentially correct. For instance, this is at least as difficult as factoring semiprimes, because it lets us compute $1+p+q+pq$ given a semiprime $pq$, and from $p+q,pq$ it's easy to compute $p,q$.
$endgroup$
– Wojowu
Apr 19 at 13:51