O. Fraser and B. Gordon, On representing a square as the sum of three squares, Amer. Math. Monthly, 76 (1969), 922-923.
这篇文章有人能找到吗？
找到一个文献On the Number of Representations of the Square of an Integer as the Sum of Three Squares
一个类似的问题： A375335 :a(n) is the smallest positive integer whose square can be represented as the sum of n distinct nonzero squares in exactly n ways, or -1 if no such integer exists.
前几项是1, 1, 25, 23, 17. 目前a(5)还未知，即找一个整数n，使$n^2$恰好有5种方式表示为5个互不相同的正整数的平方和。
f := Length, (FreeQ[#, 0] && # == Union[#]) &]]
前100项是0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 2, 4, 3, 6, 7, 13, 7, 16, 14, 21, 21, 27, 24, 31, 35, 43, 43, 60, 51, 66, 61, 88, 83, 105, 91, 137, 116, 124, 140, 185, 143, 195, 187, 233, 197, 266, 220, 317, 283, 318, 317, 371, 331, 433, 404, 476, 450, 529, 427, 620, 543, 616, 612, 735, 611, 742, 735, 864, 803, 954, 783, 1080, 905, 1053, 1038, 1269, 1038, 1314, 1222, 1331, 1265, 1545, 1309, 1733, 1521, 1615, 1553, 1951, 1607, 2037, 1834, 2145, 2001, 2216, 1959, 2538
看这个数列的增长速度，似乎也是不存在。
现在根据前面的估计公式，只需要计算到300,000项，就可以确保10000项以内没有找到的都是无解了，由此我们可以提交到A374805中了
(16:15) gp > solve(x=10,10^20,x/8-2*x^0.75-10000)
%23 = 268974.60060487191995397568614516513661
代码很简单
#include <vector>
#include <stdio.h>
#include <math.h>
#include <iostream>
#include <gmpxx.h>
#include <map>
#include <vector>
#include <numeric>
#include <algorithm>
#define LUP 300000L
#define LUB 10000
char *notprime;
int *qlist;
int *lp;
int qc;
long c;
#define ISPRIME(x) (!notprime)
#define UNSETPRIME(x) (notprime=1)
struct PF{
int pi;
int pc;
bool operator<(const PF& pf)const{
if(pi>pf.pi)return true;//sort in reverse order
if(pi<pf.pi)return false;
return pc<pf.pc;
}
};
struct PData{
int x,y;
};
std::vector<PF> factors;
typedef std::map<long, std::vector<PData> > PMAP;
#define LIMIT 5
void mergefactors(std::vector<PF>& pf1, const std::vector<PF>& pf2)
{
std::vector<PF> out;
int i,j;
for(i=0,j=0;i<pf1.size()&&j<pf2.size();){
if(pf1.pi==pf2.pi){
PF y;
y.pi=pf1.pi;
y.pc=pf1.pc+pf2.pc;
out.push_back(y);
i++;j++;
}else if(pf1.pi>pf2.pi){
out.push_back(pf1);
i++;
}else{
out.push_back(pf2);
j++;
}
}
for(;i<pf1.size();i++){
out.push_back(pf1);
}
for(;j<pf2.size();j++){
out.push_back(pf2);
}
pf1=out;
}
void getfactors(long x, std::vector<PF>& pf)
{
int pc=0;
int curp=-1;
if(x==0)return;
while(lp>=0){
if(lp==curp){
pc++;
}else{
if(curp>=0){
PF y;
y.pi=curp;
y.pc=pc;
pf.push_back(y);
}
pc=1;
curp=lp;
}
x/=qlist];
}
if(pc>0){
PF y;
y.pi=curp;
y.pc=pc;
pf.push_back(y);
}
}
void initprime()
{
qlist=(int *)malloc(2*LUP*sizeof(int));
lp=(int *)malloc(2*LUP*sizeof(int));
notprime=(char *)malloc(2*LUP*sizeof(char));
memset(notprime,0,2*LUP*sizeof(char));
UNSETPRIME(0);
UNSETPRIME(1);
long i,j;
for(i=2;i<2*LUP;i++){
if(!ISPRIME(i))continue;
{
qlist=i;
}
for(j=i*i;j<2*LUP;j+=i){
UNSETPRIME(j);
}
}
for(i=0;i<2*LUP;i++)lp=-1;
for(i=0;i<qc;i++){
for(j=qlist;j<2*LUP;j+=qlist){
lp=i;
}
}
for(i=2;i<2*LUP;i++){
getfactors(i,factors);
}
}
long getsc(long c, long d)
{
int i;
bool is_sqr=1;
std::vector<PF> pf1, pf2;
pf1=factors;
mergefactors(pf1,factors);
long r=1;
for(i=0;i<pf1.size();i++){
long p=qlist.pi];
if((pf1.pc&1)==1)is_sqr=0;
if(p%4==3&&(pf1.pc&1)==1)return 0;
if(p%4==1){
r*=pf1.pc+1;
}
}
if(is_sqr)r--;
return r;
}
long gete2(long x, long y)
{
std::vector<PF> p1=factors;
std::vector<PF>& p2=factors;
mergefactors(p1,p2);
long prod=1;
for(int i=0;i<p1.size();i++){
long pm=qlist.pi];
if((pm&1)==0)continue;
if(pm%4==3){
if(p1.pc&1)return 0;
}else{
prod*=p1.pc+1;
}
}
return prod;
}
long gete(long x)
{
std::vector<PF>& fac=factors;
long prod=1;
for(int i=0;i<fac.size();i++){
long pm = qlist.pi];
if((pm&1)==0)continue;
if(pm%4==3){
if(fac.pc&1)return 0;
}else{
prod*=fac.pc+1;
}
}
return prod;
}
std::map<long ,long > rmap;
long getr(long d)
{
std::vector<PF>& fac=factors;
long prod=1;
int i;
for(i=0;i<fac.size();i++){
long p=qlist.pi];
if((p&1)==0)continue;
int h;
long pm=1;
for(h=0;h<fac.pc;h++)pm*=p;
if(p%4==3){
prod*=pm+2*(pm-1)/(p-1);
}else{
prod*=pm;
}
}
return prod;
}
long gets(long d)
{
std::vector<PF>& fac=factors;
long prod=1;
for(int i=0;i<fac.size();i++){
long p=qlist.pi];
int m=p%8;
if(m!=1&&m!=3)continue;
prod*=2*fac.pc+1;
}
return prod;
}
long gett(long d)
{
std::vector<PF>& fac=factors;
long prod=1;
for(int i=0;i<fac.size();i++){
long p=qlist.pi];
int m=p%4;
if(m!=1)continue;
prod*=2*fac.pc+1;
}
return prod;
}
int main()
{
long d;
initprime();
memset(c,-1,sizeof(c));
for(d=1;d<LUP;d+=2){
long r=getr(d);
long s=gets(d);
long t=gett(d);
long f=(r-2*s-2*t+3)/8;
if(f<LUB&&c<0){
c=d;
}
}
for(d=1;d<LUB;d++){
printf("%ld %ld\n",d,c);
}
return 0;
}
mathe 发表于 2025-10-10 15:27
O. Fraser and B. Gordon, On representing a square as the sum of three squares, Amer. Math. Monthly,...
下载下来了.
这个公式很强：(a² + b² + c² + d²)² = (a² + b² - c² - d²)² + (2ac + 2bd)² + (2ad - 2bc)²
这个就是欧拉四平方数公式
\((a^2+b^2+c^2+d^2)(x^2+y^2+z^2+w^2)=(ax+by+cz+dw)^2+(ay-bx+cw-dz)^2+(az-bw-cx+dy)^2+(aw+bz-cy-dx)^2\)
其中取\(x=a,y=b,z=-c,w=-d\)