CF707C Pythagorean Triples

Solution Sketch

Using Euclid's formula

$$\begin{align*} a& = p^2 - q^2 \tag 1\\ b& = 2pq \tag 2\\ c& = p^2 + q^2 \tag 3\\ \end{align*}$$

There are two special cases, $n = 1$ and $n = 2$, where the right triangle can’t be formed (plug in the numbers and you will see why).

After eliminating special cases, we just need to deal with $n$ according to parity.

If $n$ is odd, we can use $\text{equation $(1)$}$ and property
$$
\begin{align}
(k + 1)^2 - k^2 &= 2k + 1 &&\text{the difference between two consecutive numbers’ square is odd}\tag 4
\end{align}
$$
We can get $k = \frac{n - 1}{2}$ (because $2k + 1 = n$), and then substitute $k$ with $\frac{n - 1}{2}$ in $\text{equation $(4)$}$ to obtain $q = \frac{n - 1}{2} + 1$ and $p = \frac{n - 1}{2}$. We can get the answers for odd from $\text{equation $(2)$ and equation $(3)$}$, which are $$2 (\frac{n - 1}{2} + 1) (\frac{n - 1}{2}) \;\text{and}\; (\frac{n - 1}{2} + 1)^2 + (\frac{n - 1}{2})^2$$

If $n$ is even, we can use $\text{equation $(2)$}$ and let $q = 1$ (then $p = \frac n2$). We can the get the answers for even from $\text{equation $(1)$ and equation $(3)$}$, which are $$(\frac{n}{2})^2 - 1 \;\text{and}\; (\frac{n}{2})^2 + 1$$

AC Code

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int main()
{
	ll n;
	scanf("%lld", &n);
	if(n <= 2)
		printf("-1\n");
	else {
		if(n & 1) {
			n = (n - 1) / 2;
			printf("%lld %lld\n", 2 * (n + 1) * n, (n + 1) * (n + 1) + n * n);
		} else {
			n /= 2;
			printf("%lld %lld\n", n * n - 1, n * n + 1);
		}
	}
	return 0;
}