A349324 Numbers whose square can be represented in exactly six ways as the sum of a positive square and a positive fourth power.

15697403475, 62789613900, 141276631275, 251158455600, 392435086875, 565106525100, 769172770275, 1004633822400, 1271489681475, 1569740347500, 1899385820475, 2260426100400, 2652861187275, 3076691081100, 3531915781875, 4018535289600, 4536549604275, 5085958725900

Offset: 1

Author

Jon E. Schoenfield, Nov 14 2021

Keywords

• nonn

Comments

Numbers z such that there are exactly 6 solutions to z^2 = x^2 + y^4 with x, y, z positive integers.

See also A348655.

Note that y^4 = z^2 - x^2 = (z - x)*(z + x), so each solution corresponds to a positive integer y and a factorization of y^4 into two positive integers f = z - x and g = z + x. We need f < g so that x > 0, and we need f == g (mod 2) so that (f + g)/2 = (z - x + z - x)/2 = z will be an integer.

Note also that it follows from y^4 = z^2 - x^2 that y < sqrt(z).

Thus, given some value zMax, we can generate the value of z for each solution to z^2 = x^2 + y^4 over the positive integers with z <= zMax by simply computing z = (f + y^4/f)/2 for (1) every divisor f (< y^2) of y^4 for each odd y < sqrt(zMax) and (2) every even divisor f (< y^2) of y^4 such that y^4/f is also even for each even y < sqrt(zMax), and discarding those results that exceed zMax. For any given y, the smaller the value of f, the larger the resulting z. So, for each y, we can test the divisors of y^4 that are less than y^2 in descending order, and move on to the next value of y as soon as we either exhaust all the allowed divisors or get a value of z that exceeds zMax.

For example, to get the z value for every solution with z <= 100, we can factor y^4 for each y in 1..9 as follows:

.

y y^4 f g = y^4/f z = (f+g)/2 (or comments)

- ---- -- --------- -------------------------------

1 1 - - -

2 16 2 8 ( 2 + 8)/2 = 5 (solution)

3 81 3 27 ( 3 + 27)/2 = 15 (solution)

" " 1 81 ( 1 + 81)/2 = 41 (solution)

4 256 8 32 ( 8 + 32)/2 = 20 (solution)

" " 4 64 ( 4 + 64)/2 = 34 (solution)

" " 2 128 ( 2 + 128)/2 = 65 (solution)

5 625 5 125 ( 5 + 125)/2 = 65 (solution)

" " 1 625 ( 1 + 625)/2 = 313 > 100

6 1296 24 54 (24 + 54)/2 = 39 (solution)

" " 18 72 (18 + 72)/2 = 45 (solution)

" " 12 108 (12 + 108)/2 = 60 (solution)

" " 8 162 ( 8 + 162)/2 = 85 (solution)

" " 6 216 ( 6 + 216)/2 = 111 > 100

7 2401 7 343 ( 7 + 343)/2 = 175 > 100

8 4096 32 128 (32 + 128)/2 = 80 (solution)

" " 16 256 (16 + 256)/2 = 136 > 100

9 6561 27 243 (27 + 243)/2 = 135 > 100

.

so, sorted in ascending order, the z values <= 100 that occur are 5, 15, 20, 34, 39, 41, 45, 65, 65, 60, 80, 85. (The smallest z value, 5, occurs only once, so it's A345645(1); 65, the smallest value that occurs twice, is A345700(1).)

From Karl-Heinz Hofmann, Nov 15 2021: (Start)

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

Examples

			15697403475^2 = 13780596525^2 +  86700^4
              = 13195420077^2 +  92208^4
              = 11468350875^2 + 103530^4
              = 10710751443^2 + 107124^4
              =  9221086875^2 + 112710^4
              =  4878327597^2 + 122148^4.
		

Links

• Jon E. Schoenfield, Table of n, a(n) for n = 1..424 (all terms < A346115(7) = 2474052064291275)
• Karl-Heinz Hofmann, A 3D animation of the terms and the least one with 7 solutions

Crossrefs

Cf. A111925, A271576, A345645 (in exactly 1 way), A345700 (in exactly 2 ways), A345968 (in exactly 3 ways), A346110 (in exactly 4 ways), A348655 (in exactly 5 ways), A346115 (the least solutions).

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