uva 10512

測試數據非常強大，一兩個錯誤就猛吃WA

利用公式解可以解出這道題目，只要利用那兩條等式換來換去就可以得到ay4+by2+c=0這個等式

再套公式即可。

//====================================================================||
// Name        : A Day in Math-land.cpp                                              ||
// Date : 2013/7/18 下午11:20:06                                               ||
// Author : GCA                                                       ||
//                  6AE7EE02212D47DAD26C32C0FE829006                  ||
//====================================================================||
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <climits>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <cctype>
#include <utility>
using namespace std;
#ifdef ONLINE_JUDGE
#define ll "%lld"
#else
#define ll "%I64d"
#endif
typedef unsigned int uint;
typedef long long int Int;
#define Set(a,s) memset(a,s,sizeof(a))
#define Write(w) freopen(w,"w",stdout)
#define Read(r) freopen(r,"r",stdin)
#define Pln() printf("\n")
#define I_de(x,n)for(int i=0;i<n;i++)printf("%d ",x[i]);Pln()
#define De(x)printf(#x"%d\n",x)
#define For(i,x)for(int i=0;i<x;i++)
#define CON(x,y) x##y
#define Pmz(dp,nx,ny)for(int hty=0;hty<ny;hty++){for(int htx=0;htx<nx;htx++){\
    printf("%d ",dp[htx][hty]);}Pln();}
#define M 10001
#define PII pair<int,int>
#define PB push_back
#define oo (1LL<<62)
#define Set_oo 0x3f
#define Is_debug true
#define debug(...) if(Is_debug)printf("DEBUG: "),printf(__VA_ARGS__)
#define FOR(it,c) for(__typeof((c).begin()) it=(c).begin();it!=(c).end();it++)
#define eps 1e-6
bool xdy(double x,double y){return x>y+eps;}
bool xddy(double x,double y){return x>y-eps;}
bool xcy(double x,double y){return x<y-eps;}
bool xcdy(double x,double y){return x<y+eps;}
int min3(int x,int y,int z){
    int tmp=min(x,y);
    return min(tmp,z);
}
int max3(int x,int y,int z){
    int tmp=max(x,y);
    return max(tmp,z);
}
Int p,q;
inline Int sqr(Int x){
    return x*x;
}
inline bool isqrt(Int x,Int &ans){
    if(x<0)return false;
    ans=sqrt(x);
    return ans*ans==x;
}
void solve(){
    Int d=sqr(p)+6LL*p*q+sqr(q);
    Int y,x,tmp,t,ansx=oo,ansy;
    bool is=isqrt(d,t);
    if(!is){
        printf("Impossible.\n");
        return ;
    }
//    debug("%lld\n",t);
    if((3LL*p+q+t)%4==0){
        tmp=(3LL*p+q+t)>>2LL;
        is=isqrt(tmp,y);
        if(is){
            bool fg=false;
            if(y&&p%y==0){
                x=p/y-y;
                fg=true;
            }else if(y==0){
                fg=isqrt(q,x);
            }
            if(x>=y&&x<ansx&&fg){
                ansx=x;
                ansy=y;
            }
            fg=false;
            y=-y;
            if(y&&p%y==0){
                x=p/y-y;
                fg=true;
            }else if(y==0){
                fg=isqrt(q,x);
            }
            if(x>=y&&x<ansx&&fg){
                ansx=x;
                ansy=y;
            }
        }
    }
    if((3LL*p+q-t)%4==0){
        tmp=(3LL*p+q-t)>>2LL;
        is=isqrt(tmp,y);
        if(is){
            bool fg=false;
            if(y&&p%y==0){
                x=p/y-y;
                fg=true;
            }else if(y==0){
                fg=isqrt(q,x);
            }
            if(x>=y&&x<ansx&&fg){
                ansx=x;
                ansy=y;
            }
            fg=false;
            y=-y;
            if(y&&p%y==0){
                x=p/y-y;
                fg=true;
            }else if(y==0){
                fg=isqrt(q,x);
            }
            if(x>=y&&x<ansx&&fg){
                ansx=x;
                ansy=y;
            }
        }
    }
    if(ansx==oo)printf("Impossible.\n");
    else printf("%lld %lld\n",ansx,ansy);
}
int main() {
    ios_base::sync_with_stdio(0);
    int test;
    scanf("%d",&test);
    int ff=0;
    while(test--){
        scanf("%lld %lld",&p,&q);
        printf("Case %d:\n",++ff);
        solve();

    }
















}