cp's OEIS Frontend

Number of non-congruent solutions of x^2 + y^2 -1 == 2xy == 0 (mod n).

This sequence combines A329586 (number of representative solutions of a^2 - (b^2 + 1) == 0 (mod m) and 2*a*b == 0 (mod m) with a*b = 0), and those from A329589 (number of representative solutions of these two congruences but with a*b not 0). - Wolfdieter Lang, Dec 14 2019

In A226746 the positive n numbers with more than two representative solutions of the congruence z^2 = +1 (mod n) are given. This is therefore a proper subsequence of the present one. - Wolfdieter Lang, Dec 14 2019

The length of row n is given in 2*A329588(n).

This table should be considered together with A329585 which gives the solutions of this congruence with a*b = 0 (real or pure imaginary solutions) for each m >= 2. Only for m = 1 is a = b, namely (a, b) = (0, 0).

Because z^2 = (a^2 - b^2) + 2*a*b*i, this is equivalent to the two congruences (i) a^2 - b^2 == 1 (mod m) and (ii) 2*a*b == 0 (mod m). Here with nonvanishing a*b.

Because with each solution z = a + b*i, with representative a and b from {1, 2, ..., m-1}, also -z = -(a + b*i), zbar:= a - b*i and -zbar = -a + b*i are solutions modulo m.

This table T(n, k) lists all pairs (a,b) for those even m = m(n) which have solutions with nonvanishing a*b.

The solutions for odd numbers start with m = 15, 35, 39, 45, 51, 55, ... . See A329589 which is a proper subsequence of A257591. For example, m = 63 is not present in A329589.

There is a symmetry (a, b) <-> (m-a, m-b) == (-a, -b) (mod m) around the middle of the pairs in each row. Such pairs correspond to z and -z (mod m). Because of this symmetry one considers first a > b (a = b cannot occur for m >= 2, see above), and later adds the a < b solutions.

Obvious solutions for each solvable even m are for m/2 odd (m/2+1, m/2) with companion (m/2-1, m/2), and in addition, if m/(2^e2) is odd, for e2 >= 2, (m-1, m/2) with companion (1, m/2) (for m = 4 this coincides with the first case).

For (positive) even m, m = 2*M, there is the following connection to Pythagorean triples (X, Y, Z). We say triples, not triangles, because X may also become negative. Eq.(i) from above becomes (i') a^2 - (b^2 + 1) = qhat*2*M, with integer qhat. This implies that a and b have opposite parity, i.e., (-1)^(a+b) = -1. The other eq. (ii) is now (ii') a*b = q*M, with integer q. Thus (q, qhat) = (Y/m, (X-1)/m).

Case A) Primitive Pythagorean triples (pPT). If a > b and gcd(a, b) = 1 then the conditions for primitive Pythagorean triples are satisfied. Set 0 < X = a^2 - b^2 = 1 + 2*qhat*M, 0 < Y = 2*a*b = 2*q*M, 0 < Z = a^2 + b^2 = r^2 (Y is even, X is odd, and Z is odd). One could interchange the role of X and Y (for triangles a catheti exchange). Note that the radius r is in general not an integer. E.g., for m=4 (a, b) = (3, 2) has r = sqrt(13), (q, qhat) = (3, 1), pPT = (5, 12, 13).

Note that the companion triple for a < b (z -> -z) has negative X. In this example the companion of (a, b) = (3, 2) is (-a, -b) (mod 4) == (4-3, 4-2) = (1, 2), and the companion pPT = (-3, 4, 5).

Case B) m even, a > b, (-1)^(a+b) = -1, gcd(a, b) = g >= 2, leads to imprimitive Pythagorean triples (ipPT) for solutions of (i') and (ii'). The first example appears for m = 20, M = 10, (a, b) = (15, 12), g = 3, (q, qhat) = (18, 4), r = 3*sqrt(41), ipPT = (81, 360, 369) = (3^2)*(9, 40, 41). The companion has (a, b) = (5, 8), which is primitive, (q, qhat) = (4, -2), with pPT = (-39, 80, 89).

In A226746 the m values with more than two representative solutions of z^2 == +1 (mod m) are given. For the corresponding solutions one has also to consider the irregular triangle A329588(n).

The number of all representative solutions z^2 == +1 (mod m), for m >= 1, is found by combining A329586 and A329588, and is given in A227091.

This sequence gives one half of the row lengths of the irregular triangle A329587.

For the number of representative solutions of this congruence for all positive moduli m and vanishing a*b see A329586.

The formula for the number of representative solutions a(n), with modulus m = m(n), given in A329587, can be found from the number of all such solutions for m = m(n) given in A227091 after subtracting the number of solutions with a*b = 0 given in A329586. For odd m = m(n) this is S(m) = 2^(r1(m) +r3(m))*(2^r1(m) - 1) - delta(r3(m), 0)*2^(r1(m)), with r1(m) and r3(m) the number of distinct primes 1 (mod 4) and 3 (mod 4) in the prime number factorization of m respectively, and delta is the Kronecker symbol. For even m this is S(m) = 2^(r1(m) + r3(m))*(2^(1+r1(m)) - 1) - delta(r3(m), 0)*2^(r1(m)) if m/2 is odd (e2 = 1), and otherwise S(m) = 2^(r2(e2(m)) + r1(m) + r3(m))*(2^(1 + r1(m) + r3(m)) - 1), with r2(e2(m)) = 1 or 2 if e2(m) = 2 or >= 3, if m/2^(e2(m)) is odd.