A345645 -id:A345645 - cp's OEIS frontend

A271576 Integers whose square is of the form x^2 + y^4, with x,y > 0 (see A111925).

5, 15, 20, 34, 39, 41, 45, 60, 65, 80, 85, 111, 125, 135, 136, 145, 150, 156, 164, 175, 180, 194, 219, 240, 245, 255, 260, 265, 306, 313, 320, 325, 340, 351, 353, 369, 371, 375, 405, 410, 444, 445, 455, 500, 505, 514, 540, 544, 580, 585, 600, 605, 609, 624, 629, 656, 671, 674, 689

Offset: 1

Michel Marcus, Apr 10 2016

nonn

If z is in this sequence, z*k^2 is, for k > 0. Also note that since there are no fourth powers in A111925, there are no squares in this sequence. - Altug Alkan, Apr 10 2016
From Karl-Heinz Hofmann, Oct 22 2021: (Start)
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. (End)
From Jon E. Schoenfield, Nov 15 2021: (Start)
Apparently, the vast majority of the terms of this sequence can be expressed as x^2 + y^4 with x,y > 0 in only one way (A345645 lists those terms), but some can be so expressed in exactly two, three, four, five, six, or more ways. Among the first 3976926961 terms of this sequence,
        3948648229 are in A345645 (exactly 1 way),
          25415062 are in A345700 (exactly 2 ways),
           2697713 are in A345968 (exactly 3 ways),
            161543 are in A346110 (exactly 4 ways),
              3989 are in A348655 (exactly 5 ways),
               424 are in A349324 (exactly 6 ways),
        and just 1 -- a(3976926961) = 2474052064291275 = A346115(7) -- is a number whose square can be written as x^2 + y^4 with x,y > 0 in exactly 7 ways. The ratios of successive counts above, i.e., 3948648229/25415062, 25415062/2697713, 2697713/161543, 161543/3989, 3989/424, and 424/1, are approximately 155.4, 9.4, 16.7, 40.5, 9.4, and 424.0. What is it that (over the interval [1, 2474052064291275], at least) makes numbers whose squares can be written as x^2 + y^4 with x,y > 0 in more than 6 ways so much rarer than those that can be written thus in exactly 6 ways? (End)

			5^2 = 25 = 9 + 16 = 3^2 + 2^4, so 5 is a term.

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

Cf. A111925, A271577, A345645 (in exactly 1 way).
Cf. A345700 (in exactly 2 ways), A345968 (in exactly 3 ways).
Cf. 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)).

Select[Range@ 200, Resolve[Exists[{x, y}, Reduce[#^2 == x^2 + y^4, {x, y}, Integers], And[x > 0, y > 0]]] &] (* Michael De Vlieger, Apr 10 2016 *)

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

is(n)=my(n2=n^2); for(b=sqrtnint(2*n-2,4)+1, sqrtint(n-1), if(issquare(n2-b^4), return(1))); 0 \\ Charles R Greathouse IV, Nov 16 2021

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

Original entry on oeis.org

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.
```

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)).

A349664 a(n) is the number of solutions for n^4 = z^2 - x^2 with {z,x} >= 1.

Original entry on oeis.org

0, 1, 2, 3, 2, 7, 2, 5, 4, 7, 2, 17, 2, 7, 12, 7, 2, 13, 2, 17, 12, 7, 2, 27, 4, 7, 6, 17, 2, 37, 2, 9, 12, 7, 12, 31, 2, 7, 12, 27, 2, 37, 2, 17, 22, 7, 2, 37, 4, 13, 12, 17, 2, 19, 12, 27, 12, 7, 2, 87, 2, 7, 22, 11, 12, 37, 2, 17, 12, 37, 2, 49, 2, 7, 22

Offset: 1

Karl-Heinz Hofmann, Dec 13 2021

nonn

If n is an odd prime^i, the number of solutions is 2*i.
If n = 2^i, the number of solutions is 2*i-1.
These two facts are not generally valid in reverse for terms > 6.
If a(n) = 2, n is an odd prime. This is generally valid in reverse.
For more information about these facts see the link.
The calculation of the terms is done with an algorithm of Jon E. Schoenfield, which is described in A349324.
Conditions to be satisfied for a valid, countable solution:
- z cannot be a square.
- z must have at least one prime factor of the form p == 1 (mod 4), a Pythagorean prime (A002144).
- If z has 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 n too.
- If z is even, x and n must be even too.
- The lower limit of the ratio x/n is sqrt(2).
- high limits of z and x:
           |    n is odd      |    n is even
  ---------+------------------+------------------
  z limit  | (n^4 + 1)/2      |  (n^4 + 4)/4
  x limit  | (n^4 + 1)/2 - 1  |  (n^4 + 4)/4 - 2

			a(6) = 7 (solutions): 6^4 = 1296 = 325^2 - 323^2 = 164^2 - 160^2 = 111^2 - 105^2 = 85^2 - 77^2 = 60^2 - 48^2 = 45^2 - 27^2 = 39^2 - 15^2.

Karl-Heinz Hofmann, Table of n, a(n) for n = 1..10000
Karl-Heinz Hofmann, What the terms can tell about n.

Cf. A000290, A000583, A271576, A349663, A002144, A002145, A346115.

Cf. A345645, A345700, A345968, A346110, A348655, A349324.

a[n_] := Length[Solve[n^4 == z^2 - x^2 && x >= 1 && z >= 1, {x, z}, Integers]]; Array[a, 75] (* Amiram Eldar, Dec 14 2021 *)

a(n) = numdiv(if(n%2, n^4, n^4/4))\2; \\ Jinyuan Wang, Dec 19 2021

cp's OEIS Frontend

A271576 Integers whose square is of the form x^2 + y^4, with x,y > 0 (see A111925).

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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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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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A349664 a(n) is the number of solutions for n^4 = z^2 - x^2 with {z,x} >= 1.

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