[CF707C]Pythagorean Triples（数学）

题目链接：http://codeforces.com/contest/707/problem/C

题意：给一个数，求和这个数匹配的勾股数。

公式题，给个链接：http://wenku.baidu.com/link?url=xKODYgXOSh20OoWMHk4stVmAxK2ibWvIOUhiDiqjGX24ALotglvjZF4aDCQ2craXk6xHvKHplN_G3JGsqr6q6_iT1aXz2S_qOOMpXaqMmBG

/*
━━━━━┒ギリギリ♂ eye！
┓┏┓┏┓┃キリキリ♂ mind！
┛┗┛┗┛┃＼○／
┓┏┓┏┓┃ /
┛┗┛┗┛┃ノ)
┓┏┓┏┓┃
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┛┗┛┗┛┃
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*/
#include <algorithm>
#include <iostream>
#include <iomanip>
#include <cstring>
#include <climits>
#include <complex>
#include <fstream>
#include <cassert>
#include <cstdio>
#include <bitset>
#include <vector>
#include <deque>
#include <queue>
#include <stack>
#include <ctime>
#include <set>
#include <map>
#include <cmath>
using namespace std;
#define fr first
#define sc second
#define cl clear
#define BUG puts("here!!!")
#define W(a) while(a--)
#define pb(a) push_back(a)
#define Rint(a) scanf("%d", &a)
#define Rll(a) scanf("%I64d", &a)
#define Rs(a) scanf("%s", a)
#define Cin(a) cin >> a
#define FRead() freopen("in", "r", stdin)
#define FWrite() freopen("out", "w", stdout)
#define Rep(i, len) for(int i = 0; i < (len); i++)
#define For(i, a, len) for(int i = (a); i < (len); i++)
#define Cls(a) memset((a), 0, sizeof(a))
#define Clr(a, x) memset((a), (x), sizeof(a))
#define Full(a) memset((a), 0x7f7f7f, sizeof(a))
#define lrt rt << 1
#define rrt rt << 1 | 1
#define pi 3.14159265359
#define RT return
#define lowbit(x) x & (-x)
#define onecnt(x) __builtin_popcount(x)
typedef long long LL;
typedef long double LD;
typedef unsigned long long ULL;
typedef pair<int, int> pii;
typedef pair<string, int> psi;
typedef pair<LL, LL> pll;
typedef map<string, int> msi;
typedef vector<int> vi;
typedef vector<LL> vl;
typedef vector<vl> vvl;
typedef vector<bool> vb;

LL a;

int main() {
    // FRead();
    while(cin >> a) {
        if(a == 1 || a == 2) {
            puts("-1");
            continue;
        }
        if(a % 2 == 0) {
            a /= 2;
            cout << a * a - 1 << " " << a * a + 1 << endl;
        }
        else {
            a /= 2;
            cout << 2 * a * a + 2 * a << " " << 2 * a * a + 2 * a + 1 << endl;
        }
    }
    RT 0;
}