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Squares are in A075577.

The companion sequence is A212354.

There are at most two incongruent solutions of this congruence due to the degree. The fact that there are precisely two such solutions for each prime of the form 4*k+1 (see A002144) is due to the reduction of this problem to one of quadratic residues, namely to X^2 == -1 (mod 2p), with p a prime (see the Nagell reference, given in A210848, pp. 132-3, especially theorem 77), adapted to the quadratic form f(x) = 2*x^2 + 2*x + 1, with discriminant D=-4. This congruence with composite modulus has exactly two incongruent solutions because X^2 == -1 (mod 2) has only the solution +1 modulo 2 (odd numbers), and X^2 == -1 (mod p) has (at least one) solution if the Legendre symbol (-1/p) = +1 (i.e., if -1 is a quadratic residue modulo p). Now (-1/p) = (-1)^(p-1)/2 (see, e.g., the Niven-Zuckerman-Montgomery reference given in A001844, Theorem 3.2 (1), p. 132). Hence there is a solution modulo p iff p == 1 (mod 4). Call the smallest positive one X0, with 0 < X0 < p-1. Then one also has the incongruent solution X1 := p-X0. This implies that there are precisely two incongruent solution of the original congruence modulo 2*p for each 1 (mod 4) prime (see, e.g., Nagell's book, pp. 83-4, Theorem 46). If u is a solution for p = A002144(n) (the existence of u has just been proved) then also the companion v := p-1-u satisfies this congruence, and v is incongruent to u modulo p.

Note that x^2 + (x+1)^2 = 4*T(x) + 1, with the triangular numbers A000217.

The primes with x^2 +(x+1)^2 = prime (necessarily from A002144) are found under A027862. The corresponding x values are found under A027861. These x values explain the positions n' where a(n') is smaller than a(n'-1) (for n'>=6): determine k with x=A027861(k), and then n' from A027862(k) = A002144(n'). Note that a(n') = x for such values n'. E.g., n'=6 with a(6)=4: x=4=A027861(3), p=41=A027862(3) = A002144(6). These values n' are n' = 1, 2, 6, 8, 14, 19, 30, ...

All positive solutions of this congruence are provided by the two sequences with entries u(n,k) = a(n) + k*A002144(n) and v(n,k) = A212354(n) + k*A002144(n), n >= 1, k >= 0. For the cases p = 5, 13 and 17 see A047219, A212160 and A212161, respectively, where the even-indexed numbers are the u(n,k) and the odd-indexed ones the v(n,k) (bisection).

2*a(n) + 1 = A206549(n), the smallest positive nontrivial solution of X^2 == +1 (Modd A002144(n)). For the next larger solution 2*A212354(n) + 1 >= p, hence it does not belong to the restricted residue system Modd A002144(n).

There are no common representations of two, three or six squares for n < 13, so

a(n) = A218208(n) + A218210(n) + A218212(n); n < 13.

See A027862 for primes of the form x^2+(x+1)^2 = 2x^2+2x+1.

See A027864 for primes of the form x^2+(x+1)^2+(x+2)^2 = 3x^2+6x+5.

It is an open question whether either of these polynomials produces an infinite number of primes. This sequence lists the values of x that produce a prime in both polynomials. x must be congruent to 0 or 2 (mod 4) and all the generated primes are of the form 4k+1.

Integers m such both m and m+1 are terms in A027861.

All corresponding primes are == 1 mod 4 (A002144 Pythagorean primes) and terms in A027862.

No such m such that also (m+2)^2 + (m+3)^2 is prime.

The corresponding n are a subsequence of A001333; see example.

The similar sequence A027861 (complement of A012132) is related to primes.

Consecutive terms are (254,255) (4099,4100) (11159,11160) (25094,25095) (31754,31755) (40189,40190) ... Conjecture: There are infinitely many consecutive values.

Among quadratic polynomials in k of the form a*k^2 + a*k - 1 the value a=70 gives the most primes for any a in the range 1<=a<=300, at least up to k=40000. Here a and k are positive integers. Other "good" values of a are a=250, a=99, and a=19.