#include <stdio.h>
#include <iostream>
#include <tchar.h>
#include <stdint.h>
#include <math.h>

using namespace std;

// Enumerates (n, A081324(n)) for 0 <= n < count.
void a081324(const uint64_t count) {
	uint64_t n = 1;      // Index in Sequence
	uint64_t a_n;        // a(n)
	uint64_t e;          // sqrt(a(n)/2)
	uint64_t iSqr, jSqr; // Tests for iSqr + jSqr = a_n.
	uint64_t i;          // sqrt(iSqr)
	uint64_t j;          // sqrt(jSqr)
	
	bool isSquareSum;    // Is 2e^2 a sum of two perfect squares i^2 and j^2 such that i != j?

	for (e = 0; n <= count; e++) {
		isSquareSum = false;
		a_n = 2 * e * e;
		// For counting the number of integer pairs (i, j) such that i^2 + j^2 == 2e^2
		// and such that i > j (to avoid repeats), i > e. Furthermore, i^2 != 2e^2, as
		// 2e^2 cannot be a perfect square, and i^2 <= 2e^2, as i^2 > 2e^2 would imply
		// that j^2 < 0, which is obviously impossible, so i^2 < 2e^2.
		for (i = e + 1, iSqr = i * i; iSqr < a_n; i++, iSqr = i * i) {
			jSqr = a_n - iSqr; // Solves for j^2.
			j = (uint64_t)(sqrtl((long double)jSqr) + 0.5); // Calculates j rounded to the nearest integer.
			if (j * j == jSqr) { // If jSqr is a perfect square
				isSquareSum = true;
				break;
			}
		}
		if (!isSquareSum) {
			cout << n << ' ' << a_n << endl;
			n++;
		}
	}
}