A345645 Numbers whose square can be represented in exactly one way as the sum of a square and a biquadrate (fourth power).

5, 15, 20, 34, 39, 41, 45, 60, 80, 85, 111, 125, 135, 136, 150, 156, 164, 175, 180, 194, 219, 240, 245, 255, 265, 306, 313, 320, 325, 340, 351, 353, 369, 371, 375, 405, 410, 444, 445, 455, 500, 505, 514, 540, 544, 600, 605, 609, 624, 629, 656, 671, 674, 689

Offset: 1

Author

Mohammad Tejabwala, Jun 21 2021

Keywords

• nonn

Comments

Numbers z such that there is exactly one solution to z^2 = x^2 + y^4.

From Karl-Heinz Hofmann, Oct 21 2021: (Start)

No term can be a square (see the comment from Altug Alkan in A111925).

Terms must have at least one prime factor of the form p == 1 (mod 4), a Pythagorean prime (A002144).

Additionally, if the terms have prime factors of the form p == 3 (mod 4), which are in A002145, then they must appear in the prime divisor sets of x and y too.

The special prime factor 2 has the same behavior, i.e., if the term is even, x and y must be even too. (End)

Examples

			3^2 + 2^4 = 9 + 16 = 25 = 5^2, so 5 is a term.
60^2 + 5^4 = 63^2 + 4^4 = 65^2, so 65 is not a term.
		

Links

• Karl-Heinz Hofmann, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000290, A000583, A180241, A271576 (all solutions).

Cf. A345700 (2 solutions), A345968 (3 solutions), A346110 (4 solutions), A348655 (5 solutions), A349324 (6 solutions), A346115 (the least solutions).

Cf. A002144 (p == 1 (mod 4)), A002145 (p == 3 (mod 4)).

Programs

Mathematica

Select[Range@100,Length@Solve[x^2+y^4==#^2&&x>0&&y>0,{x,y},Integers]==1&] (* Giorgos Kalogeropoulos, Jun 25 2021 *)

PARI

inlist(list, v) = for (i=1, #list, if (list[i]==v, return(1)));
isok(m) = {my(list = List()); for (k=1, sqrtnint(m^2, 4), if (issquare(j=m^2-k^4) && !inlist(vecsort([k^4,j^2])), listput(list, vecsort([k^4,j^2])));); #list == 1;} \\ Michel Marcus, Jun 26 2021

Python

terms = []
for i in range(1, 700):
    occur = 0
    ii = i*i
    for j in range(1, i):
        k = int((ii - j*j) ** 0.25)
        if k*k*k*k + j*j == ii:
            occur += 1
    if occur == 1:
        terms.append(i)
print(terms)