java代码：
1 import java.math.BigInteger; 2 import java.util.Scanner; 3 4 public class Main 5 { 6 public static void solve(int n) 7 { 8 BigInteger N, p1, p2, q1, q2, a0, a1, a2, g1, g2, h1, h2,p,q; 9 g1 = q2 = p1 = BigInteger.ZERO; 10 h1 = q1 = p2 = BigInteger.ONE; 11 a0 = a1 = BigInteger.valueOf((int)Math.sqrt(1.0*n)); 12 BigInteger ans=a0.multiply(a0); 13 if(ans.equals(BigInteger.valueOf(n))) 14 { 15 System.out.println("No solution!"); 16 return; 17 } 18 N = BigInteger.valueOf(n); 19 while (true) 20 { 21 g2 = a1.multiply(h1).subtract(g1); 22 h2 = N.subtract(g2.pow(2)).divide(h1); 23 a2 = g2.add(a0).divide(h2); 24 p = a1.multiply(p2).add(p1); 25 q = a1.multiply(q2).add(q1); 26 if (p.pow(2).subtract(N.multiply(q.pow(2))).compareTo(BigInteger.ONE) == 0) break; 27 g1 = g2;h1 = h2;a1 = a2; 28 p1 = p2;p2 = p; 29 q1 = q2;q2 = q; 30 } 31 System.out.println(p+" "+q); 32 } 33 34 public static void main(String[] args) 35 { 36 Scanner cin = new Scanner(System.in); 37 int t=cin.nextInt(); 38 while(t>0) 39 { 40 int n=cin.nextInt(); 41 if(n==0) 42 { 43 System.out.println("1 1"); 44 continue; 45 } 46 solve(n); 47 t--; 48 } 49 } 50 }
c++代码：
1 #include<bits/stdc++.h> 2 #define ll long long 3 #define dbg(x) cout<<#x<<" == "<<x<<endl 4 int rd(){int x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){ if(ch=='-'){f=-1;} ch=getchar();}while(ch>='0'&&ch<='9'){x=x*10+ch-'0'; ch=getchar();}return x*f;} 5 ll rdll(){ll x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){ if(ch=='-'){f=-1;} ch=getchar();}while(ch>='0'&&ch<='9'){x=x*10+ch-'0'; ch=getchar();}return x*f;} 6 using namespace std; 7 const int maxn=1e5+5; 8 const int mod=128; 9 const int Mod=13331; 10 ll a[20000]; 11 bool pell_minimum_solution(ll n,ll &x0,ll &y0){ 12 ll m=(ll)sqrt((double)n); 13 double sq=sqrt(n); 14 int i=0; 15 if(m*m==n)return false;//当n是完全平方数则佩尔方程无解 16 a[i++]=m; 17 ll b=m,c=1; 18 double tmp; 19 do{ 20 c=(n-b*b)/c; 21 tmp=(sq+b)/c; 22 a[i++]=(ll)(floor(tmp)); 23 b=a[i-1]*c-b; 24 //printf("%lld %lld %lld ",a[i-1],b,c); 25 }while(a[i-1]!=2*a[0]); 26 ll p=1,q=0; 27 for(int j=i-2;j>=0;j--){ 28 ll t=p; 29 p=q+p*a[j]; 30 q=t; 31 //printf("a[%d]=%lld %lld %lld ",j,a[j],p,q); 32 } 33 if((i-1)%2==0){x0=p;y0=q;} 34 else{x0=2*p*p+1;y0=2*p*q;} 35 return true; 36 } 37 38 int main(){ 39 ll n,x,y; 40 int t=rd(); 41 while(t--){ 42 n=rdll(); 43 if(pell_minimum_solution(n,x,y)){ 44 printf("%d %d ",x,y); 45 // printf("%lld^2-%lld*%lld^2=1 ",x,n,y); 46 // printf("%lld-%lld=1 ",x*x,n*y*y); 47 } 48 } 49 }
详解博客：https://blog.csdn.net/wh2124335/article/details/8871535
补充：
1 import java.io.*; 2 import java.math.*; 3 import java.util.*; 4 5 public class pei { 6 static long [] a = new long [1000]; 7 static BigInteger x; 8 static BigInteger y; 9 10 public static void main(String [] args) 11 { 12 Scanner cin = new Scanner(System.in); 13 int t=cin.nextInt(); 14 while(t>0){ 15 long n=cin.nextInt(); 16 if(pell_solution(n)){ 17 18 System.out.println(x+" "+y); 19 }else{ 20 System.out.print(""no solution","); 21 } 22 } 23 } 24 public static boolean pell_solution(long D){ 25 double sq=Math.sqrt((double)D); 26 long m=(long) Math.floor(sq); 27 int i=0; 28 if(m*m==D)return false; 29 a[i++]=m; 30 long b=m,c=1; 31 double tmp; 32 do{ 33 c=(D-b*b)/c; 34 tmp=(sq+b)/c; 35 a[i++]=(long)(Math.floor(tmp)); 36 b=a[i-1]*c-b; 37 }while(a[i-1]!=2*a[0]); 38 BigInteger p=new BigInteger("1"); 39 BigInteger q=new BigInteger("0"); 40 BigInteger t; 41 for(int j=i-2;j>=0;j--){ 42 t=p; 43 p=q.add(p.multiply(BigInteger.valueOf(a[j]))); 44 q=t; 45 } 46 if((i-1)%2==0){ 47 x=p;y=q; 48 }else{ 49 x=BigInteger.valueOf(2).multiply(p).multiply(p).add(BigInteger.ONE); 50 y=BigInteger.valueOf(2).multiply(p).multiply(q); 51 } 52 return true; 53 } 54 }