变形得到 $a^2 = (c + b) (c - b)$，枚举 $a^2$ 的因数即可。

#include <bits/stdc++.h>
using namespace std;
int main()
{
	long long n;
	cin >> n;
	n *= n;
	for (long long i = 1; i * i < n; i++)
		if (n % i == 0 && i % 2 == n / i % 2)
			cout << (n / i - i) / 2 << ' ' << (n / i + i) / 2 << endl;
	return 0;
}

数学题，分析如下：

设 $d = c - b$，$e = c + b$。

由性质可得：

然后枚举 $d$ 即可。

AC code：

#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define pb push_back
#define mp make_pair

int main() {
    ll a;
    cin >> a;
    vector<pair<ll, ll>> res;
    if (a != 1) {
        ll a2 = a * a;
        for (ll d = 1; d * d <= a2; ++d) {
            if (a2 % d != 0) continue;
            ll e = a2 / d;
            if ((d + e) % 2 != 0) continue;
            ll b = (e - d) / 2, c = (e + d) / 2;
            if (b > 0) {
                res.pb(mp(b, c));
            }
        }
    }
    sort(res.begin(), res.end(), [](const pair<ll, ll>& x, const pair<ll, ll>& y) {
        return x.first > y.first;
    });
    for (int i = 0; i < res.size(); ++i) {
        cout << res[i].first << ' ' << res[i].second << endl;
    }
    return 0;
}