A345700 Numbers whose square can be represented in exactly two ways as the sum of a positive square and a positive fourth power.

65, 145, 260, 580, 585, 1025, 1040, 1105, 1305, 2320, 2340, 2465, 3185, 3625, 4100, 4160, 4335, 4420, 4810, 5220, 5265, 6625, 7105, 7585, 7865, 8405, 9225, 9280, 9360, 9860, 9945, 10985, 11745, 12740, 14500, 16400, 16465, 16640, 17340, 17545, 17680, 19240

Offset: 1

Author

Karl-Heinz Hofmann, Jun 24 2021

Keywords

• nonn

Comments

Terms are numbers z such that there are exactly two solutions to z^2 = x^2 + y^4, where x, y and z belong to the set of positive integers.

Terms cannot 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).

If the terms additionally 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 lower limit of the ratio x/y is sqrt(2).

Examples

			29679^2 = 29640^2 + 39^4, so 29679 is not a term (only one solution).
17680^2 = 15088^2 + 96^4 = 17472^2 + 52^4, so 17680 is a term.
36900^2 = 36000^2 + 90^4 = 36828^2 + 48^4, so 36900 is a term.
18785^2 = 18207^2 + 68^4 = 17340^2 + 85^4 = 13983^2 + 112^4, so 18785 is not a term (three solutions).
		

Links

• Jon E. Schoenfield, Table of n, a(n) for n = 1..10000
• Karl-Heinz Hofmann, All valid {z,x1,y1,x2,y2} sets up to 10^7
• Karl-Heinz Hofmann, Python code for multiple solutions

Crossrefs

Cf. A111925, A271576, A345645 (in exactly 1 way), A345968 (in exactly 3 ways), A346110 (in exactly 4 ways), A348655 (in exactly 5 ways), A349324 (in exactly 6 ways), A346115 (the least solutions).

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

Programs

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 == 2;} \\ Michel Marcus, Jun 26 2021

PARI

is(n)=my(n2=n^2,s); for(y=sqrtnint(2*n-2,4)+1,sqrtint(n-1), if(issquare(n2-y^4) && s++>2, return(0))); s==2; \\ Charles R Greathouse IV, Jul 02 2021

Python

# see link above.