A271576 -id:A271576 - cp's OEIS frontend

A346110 Numbers whose square can be represented in exactly four ways as the sum of a positive square and a positive fourth power.

469625, 1878500, 2224625, 4226625, 7514000, 8898500, 11740625, 15289625, 16906500, 20021625, 23011625, 25716665, 30056000, 35594000, 38039625, 46962500, 54316275, 55615625, 56824625, 61158500, 67626000, 79366625, 80086500, 92046500, 92481870

Offset: 1

Karl-Heinz Hofmann, Jul 05 2021

nonn

Terms are numbers z such that there are exactly four 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 special prime factor 2 has the same behavior. Means: If the term is even, x and y must be even too.
Apparently, all terms are divisible by 65. The divided terms are in A346594. Are there exceptions for n > 25? - Hugo Pfoertner, Jul 14 2021, Jul 29 2021
Yes, there are exceptions: a(44,46,53,95,97) are not divisible by 65 (5*13) but they have in common: They are divisible by 145 (5*29). - Karl-Heinz Hofmann, Aug 28 2021

			29679^2 = 29640^2 + 39^4, so 29679 is not a term (only one solution).
60^2 + 5^4 = 63^2 + 4^4 = 65^2, so 65 is not a term (only two solutions).
572^2 + 39^4 = 1500^2 + 25^4 = 1575^2 + 20^4 = 1625^2, so 1625 is not a term (only three solutions).
165308^2 + 663^4 = 349575^2 + 560^4 = 433500^2 + 425^4 = 455175^2 + 340^4 = 469625^2, so 469625 is a term (four solutions).

Jon E. Schoenfield, Table of n, a(n) for n = 1..10000 (first 97 terms from Karl-Heinz Hofmann)
Karl-Heinz Hofmann, All valid {z,x1,y1,x2,y2,x3,y3,x4,y4} sets up to 10^9.
Karl-Heinz Hofmann, A 3D Animation of the solutions up to 10^9.
Karl-Heinz Hofmann, Python code (not only for 4 Solutions).

Cf. A000290, A000583.
Cf. A271576 (all solutions), A345645 (one solution), A345700 (two solutions), A345968 (three solutions), A348655 (five solutions), A349324 (six solutions), A346115 (the least solutions).
Cf. A002144 (p == 1 (mod 4)), A002145 (p == 3 (mod 4)).

Python
```
# See Hofmann link.
```

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

Original entry on oeis.org

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

Mohammad Tejabwala, Jun 21 2021

nonn

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)

			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.

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

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)).

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

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

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)

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

Original entry on oeis.org

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

Karl-Heinz Hofmann, Jun 24 2021

nonn

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).

			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).

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

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)).

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

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.
```

A345968 Numbers whose square can be represented in exactly three ways as the sum of a positive square and a positive fourth power.

Original entry on oeis.org

1625, 6500, 14625, 18785, 24505, 26000, 40625, 58500, 75140, 79625, 88985, 98020, 104000, 120250, 131625, 162500, 169065, 196625, 220545, 234000, 274625, 296225, 300560, 318500, 355940, 365625, 392080, 416000, 481000, 526500, 547230, 586625, 611585, 612625

Offset: 1

Karl-Heinz Hofmann, Jun 30 2021

nonn

Terms are numbers z such that there are exactly 3 solutions to z^2 = x^2 + y^4, where x, y and z belong to the set of positive integers.
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.

			29640^2 + 39^4 = 29679^2; 29679 is not a term (only 1 solution).
60^2 + 5^4 = 63^2 + 4^4 = 65^2; 65 is not a term (only 2 solutions).
572^2 + 39^4 = 1500^2 + 25^4 = 1575^2 + 20^4 = 1625^2; 1625 is a term (3 solutions).
165308^2 + 663^4 = 349575^2 + 560^4 = 433500^2 + 425^4 = 455175^2 + 340^4 = 469625^2; 469625 is not a term (4 solutions).

Jon E. Schoenfield, Table of n, a(n) for n = 1..10000
Karl-Heinz Hofmann, All valid {z,x1,y1,x2,y2,x3,y3} sets up to 10^8

Cf. A271576 (1 and more solutions), A345645 (1 solution), A345700 (2 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)).

A346115 Least number k such that k^2 can be expressed in exactly n ways as x^2 + y^4 with {x, y} >= 1.

Original entry on oeis.org

5, 65, 1625, 469625, 642916625, 15697403475, 2474052064291275

Offset: 1

Karl-Heinz Hofmann, Jul 05 2021

a(5) <= 642916625. - Hugo Pfoertner, Jul 07 2021
From Karl-Heinz Hofmann, Sep 02 2021: (Start)
a(6) <= 15697403475.
2 * 10^10 < a(7) <= 2474052064291275.
These two conjectured values arise from the "green group". Up to term a(5) the least solutions are in the "blue group". Follow the links below to get more information about the different colored groups.
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)

			a(1)=A345645(1); a(2)=A345700(1); a(3)=A345968(1); a(4)=A346110(1).

Karl-Heinz Hofmann, All valid {z,x1,y1,x2,y2,x3,y3,x4,y4} sets up to 10^9
Karl-Heinz Hofmann, A 3D Animation of the 4 solutions and the least 5 solution up to 10^9

Cf. A000290, A000583, A111925.
Cf. A271576 (1 or more solutions), A345645 (1 solution),  A345700 (2 solutions), A345968 (3 solutions), A346110 (4 solutions), A348655 (5 solutions), A349324 (6 solutions).
Cf. A002144 (p == 1 (mod 4)), A002145 (p == 3 (mod 4)).

a(5) confirmed by Martin Ehrenstein, Jul 09 2021
a(6) confirmed by Karl-Heinz Hofmann, Oct 15 2021
a(7) confirmed by Jon E. Schoenfield, Nov 15 2021

A348655 Numbers whose square can be represented in exactly five ways as the sum of a positive square and a positive fourth power.

Original entry on oeis.org

642916625, 2571666500, 4418701625, 5786249625, 10286666000, 16072915625, 17674806500, 20931496625, 23144998500, 31502914625, 39768314625, 41146664000, 52076246625, 57801168750, 64291662500, 70699226000, 77792911625, 83725986500, 92579994000, 108652909625

Offset: 1

Karl-Heinz Hofmann, Oct 27 2021

nonn

Numbers z such that there are exactly 5 solutions to z^2 = x^2 + y^4.
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.
Some other terms of the sequence: 20931496625, 23144998500, 31502914625, 41146664000, 52076246625, 64291662500, 77792911625, 83725986500, 92579994000, 108652909625, 126011658500, 144656240625, 164586656000. - Chai Wah Wu, Oct 29 2021

			5786249625^2 = 5785404300^2 +  9945^4
5786249625^2 = 5608211175^2 + 37740^4
5786249625^2 = 5341153500^2 + 47175^4
5786249625^2 = 4307113575^2 + 62160^4
5786249625^2 = 2036759868^2 + 73593^4

Jon E. Schoenfield, Table of n, a(n) for n = 1..3989 (all terms < A346115(7) = 2474052064291275)

Cf. A111925, A271576, A345645 (in exactly 1 way), A345700 (in exactly 2 ways), A345968 (in exactly 3 ways), A346110 (in exactly 4 ways), A349324 (in exactly 6 ways), A346115 (the least solutions).

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

a(8) and beyond from Jon E. Schoenfield, Nov 14 2021

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

Original entry on oeis.org

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

Offset: 1

Jon E. Schoenfield, Nov 14 2021

nonn

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)

			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.

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

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)).

A293283 Numbers n such that n^2 = a^2 + b^5 for positive integers a b and n.

Original entry on oeis.org

6, 9, 18, 40, 42, 68, 75, 90, 99, 105, 122, 126, 130, 174, 192, 196, 225, 251, 257, 288, 315, 325, 330, 350, 405, 490, 499, 504, 516, 528, 546, 550, 576, 614, 651, 665, 684, 726, 735, 744, 849, 882, 900, 920, 936, 974, 1025, 1032, 1036, 1107, 1140, 1183, 1200

Offset: 1

XU Pingya, Oct 04 2017

nonn

For n > 0, k = (n + 1)(2n + 1)^2 is a term in this sequence, because k^2 = (n * (2n + 1)^2)^2 + (2n + 1)^5. Examples: 18, 75, 196, 405, 726, 1183.
When z^2 = x^2 + y^2 (i.e., z = A009003(n)), (z * y^4)^2 = (x * y^4)^2 + (y^2)^5. Thus z * y^4 is a term in this sequence. For example, 1200. More generally, for positive integer i, j and k, x^(5i - 5) * y^(5j - 1) * z^(5k - 5) is in this sequence.
When z^2 = x^2 + y^3 (i.e., z = A070745(n)), (z * y)^2 = (x * y)^2 + y^5. Thus z * y is in this sequence. E.g. 6, 18, 40, ... . More generally, for positive integer i, j and k, x^(5i - 5) * y^(5j - 4) * z^(5k - 4) is in this sequence.
When z^2 = x^2 + y^4 (i.e., z = A271576(n)), (z * y^3)^2 = (x * y^3)^2 + (y^2)^5. Thus z * y^3 is also in this sequence. E.g. 40, 405, 1107, ... . More generally, for positive integer i, j and k, x^(5i - 5) * y^(5j - 2) * z^(5k - 4) is in this sequence.

			6^2 = 2^2 + 2^5.
9^2 = 7^2 + 2^5.

Chai Wah Wu, Table of n, a(n) for n = 1..10000

Cf. A009003, A070067, A070745, A228946, A271576, A274035.

c[n_]: = Count[n^2 - Range[(n^2 - 1)^(1/5)]^5, _?(IntegerQ[Sqrt[#]] &)] > 0;
Select[Range[1200], c]

isok(n) = for (k=1, n-1, if (ispower(n^2-k^2, 5), return (1));); return (0); \\ Michel Marcus, Oct 06 2017

A293694 Numbers z such that x^2 + y^8 = z^2 for positive integers x and y.

Original entry on oeis.org

20, 34, 65, 135, 320, 369, 544, 1040, 1095, 1305, 1350, 1404, 1620, 1625, 1746, 1971, 2056, 2160, 2379, 2754, 3060, 3281, 3996, 4100, 4470, 5120, 5265, 5904, 6625, 7825, 7830, 8194, 8575, 8704, 8796, 10250, 10935, 11125, 11700, 12500, 13154, 14500, 15579

Offset: 1

XU Pingya, Oct 16 2017

nonn

Let i, j and k be nonnegative integers, m > n be positive integers. As ((m^2 - n^2)^(4*i+1) * (2*m*n)^(4*j+3) * (m^2 + n^2)^(4*k))^2 + ((m^2 - n^2)^i * (2*m*n)^(j+1) * (m^2 + n^2)^k)^8 = ((m^2 - n^2)^(4*i) * (2*m*n)^(4*j+3) * (m^2 + n^2)^(4*k+1))^2, so that the number of the form (m^2 - n^2)^(4*i) * (2*m*n)^(4*j+3) * (m^2 + n^2)^(4*k+1) is a term.
When (x, y, z) is a solution of x^2 + y^4 = z^2 (i.e., z = A271576(n)), (x^(4*i+1) * y^(4*j+2) * z^(4*k), x^i * y^(j+1) * z^k, x^(4*i) * y^(4*j+2) * z^(4*k+1)) is a solution of x^2 + y^8 = z^2.
When (x, y, z) is a solution of x^2 + y^6 = z^2 (i.e., z = A293690(n)), (x^(4*i+1) * y^(4*j+1) * z^(4*k), x^i * y^(j+1) * z^k, x^(4*i) * y^(4*j+1) * z^(4*k+1)) is a solution of x^2 + y^8 = z^2.
When (x, y, z) is a solution of x^2 + y^8 = z^2, (x^(4*i+1) * y^(4*j) * z^(4*k), x^i * y^(j+1) * z^k, x^(4*i) * y^(4*j) * z^(4*k+1)) is also a solution of x^2 + y^8 = z^2.

			12^2 + 2^8 = 20^2, 20 is a term.
63^2 + 2^8 = 65^2, 65 is a term.

Karl-Heinz Hofmann, Table of n, a(n) for n = 1..13695

Cf. A009003, A228946, A271576, A293283, A293690, A293692.

z[n_] := Count[n^2 - Range[(n^2 - 1)^(1/8)]^8, _?(IntegerQ[Sqrt[#]] &)] > 0; Select[Range[16000], z]

A271577 Powers, squares excepted, of the form x^2+y^4, with x,y > 0 (see A111925).

Original entry on oeis.org

32, 512, 2197, 3125, 8000, 8192, 15625, 64000, 100000, 131072, 1419857, 1953125, 2097152, 3200000, 7189057, 8000000, 8998912, 9765625, 11390625, 11881376, 30664297, 32768000, 33554432, 35287552, 62748517, 64000000, 69343957, 115856201, 125000000, 221445125, 262144000, 272097792

Offset: 1

Michel Marcus, Apr 10 2016

nonn

			32 = 2^5 = 16 + 16 = 2^2 + 2^4, so 32 is a term.

Chai Wah Wu, Table of n, a(n) for n = 1..1255

Cf. A111925, A271576.

isok(n) = {if ((p=ispower(n)) && (p != 2), for (k=1, n-1, if (issquare(k) && (p=ispower(n-k, 4)), return (1));););}

cp's OEIS Frontend

A346110 Numbers whose square can be represented in exactly four ways as the sum of a positive square and a positive fourth power.

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

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

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A345968 Numbers whose square can be represented in exactly three ways as the sum of a positive square and a positive fourth power.

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A346115 Least number k such that k^2 can be expressed in exactly n ways as x^2 + y^4 with {x, y} >= 1.

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A348655 Numbers whose square can be represented in exactly five ways as the sum of a positive square and a positive fourth power.

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

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A293283 Numbers n such that n^2 = a^2 + b^5 for positive integers a b and n.

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