Let n be a positive integer. Then the number of common factors of n2 + 3n + 1 and n2 + 4n + 3 is
Let x and y be consecutive integers. Suppose m be the number of solutions of x3−y3=3k2 and n be the number of solutions of x3−y3=2t2 , where k, t are integers. Then m + n equals:
∑πm−1tan−1(2mm4+m2+2) is equal to