Fermat's last theorem (number theory) ------ HDU 6441 Find Integer

Find Integer

people in USSS love math very much, and there is a famous math problem .

give you two integers $n$, $a$,you are required to find $2$ integers $b$, $c$ such that $a^{n}+b^{n}=c^{n}$.

Input

one line contains one integer $T$ ; $(1 \ leq T \ leq 1000000)$

next $T$ lines contains two integers $n$, $a$; ( $0 \ leq n \ leq 1000000000,3 \ leq a \ leq 40000$)

Output

print two integers $b$, $c$ if $b$, $c$ exits; ( $1 \leq b,c \leq 1000000$);
else print two integers -1 -1 instead.

Sample Input

1
2 3

Sample Output

4 5

answer

$a^{n}+b^{n}=c^{n}$ , gives $n,a$ , obtained $b,c$
When $n \ geq 3 || n = 0$ by Fermat's last theorem shows that no solution.
when $n = 1$ , the take $b = 1,c = a + 1$ to
when $n = 2$ , the
if $a$ is an odd number, $b = \frac{a^{2}-1}{2},c = \frac{a^{2}+1}{2}$
If the $a$ is an even number, $b = \frac{a^{2}}{4}-1,c = \frac{a^{2}}{4}+1$

C ++ code

#include <iostream>

using namespace std;

void solve(int n,int a){
    if(n > 2 || n == 0){
        printf("%d %d\n",-1,-1);
    }else if(n == 1){
        printf("%d %d\n",1,a + 1);
    } else if(n == 2){
        if(a % 2 == 1)
            printf("%d %d\n",(a * a - 1)/2,(a * a + 1)/2);
        else {
            if (a == 2)
                printf("%d %d\n",-1,-1);
            else
                printf("%d %d\n", (a * a / 4) - 1, (a * a / 4) + 1);
        }
    }
}

int main(){

    int t;
    cin >>t;
    for(int i = 0;i < t;i++){
        int n,a;
        scanf("%d%d",&n,&a);
        solve(n,a);
    }

    return 0;
}