cp's OEIS Frontend

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

%I A349324 #25 Dec 06 2021 08:37:15
%S A349324 15697403475,62789613900,141276631275,251158455600,392435086875,
%T A349324 565106525100,769172770275,1004633822400,1271489681475,1569740347500,
%U A349324 1899385820475,2260426100400,2652861187275,3076691081100,3531915781875,4018535289600,4536549604275,5085958725900
%N A349324 Numbers whose square can be represented in exactly six ways as the sum of a positive square and a positive fourth power.
%C A349324 Numbers z such that there are exactly 6 solutions to z^2 = x^2 + y^4 with x, y, z positive integers.
%C A349324 See also A348655.
%C A349324 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.
%C A349324 Note also that it follows from y^4 = z^2 - x^2 that y < sqrt(z).
%C A349324 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.
%C A349324 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:
%C A349324 .
%C A349324   y   y^4   f  g = y^4/f     z = (f+g)/2 (or comments)
%C A349324   -  ----  --  ---------  -------------------------------
%C A349324   1     1   -       -                        -
%C A349324   2    16   2       8     ( 2 +   8)/2  =    5 (solution)
%C A349324   3    81   3      27     ( 3 +  27)/2  =   15 (solution)
%C A349324   "    "    1      81     ( 1 +  81)/2  =   41 (solution)
%C A349324   4   256   8      32     ( 8 +  32)/2  =   20 (solution)
%C A349324   "    "    4      64     ( 4 +  64)/2  =   34 (solution)
%C A349324   "    "    2     128     ( 2 + 128)/2  =   65 (solution)
%C A349324   5   625   5     125     ( 5 + 125)/2  =   65 (solution)
%C A349324   "    "    1     625     ( 1 + 625)/2  =  313 > 100
%C A349324   6  1296  24      54     (24 +  54)/2  =   39 (solution)
%C A349324   "    "   18      72     (18 +  72)/2  =   45 (solution)
%C A349324   "    "   12     108     (12 + 108)/2  =   60 (solution)
%C A349324   "    "    8     162     ( 8 + 162)/2  =   85 (solution)
%C A349324   "    "    6     216     ( 6 + 216)/2  =  111 > 100
%C A349324   7  2401   7     343     ( 7 + 343)/2  =  175 > 100
%C A349324   8  4096  32     128     (32 + 128)/2  =   80 (solution)
%C A349324   "    "   16     256     (16 + 256)/2  =  136 > 100
%C A349324   9  6561  27     243     (27 + 243)/2  =  135 > 100
%C A349324 .
%C A349324 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).)
%C A349324 From _Karl-Heinz Hofmann_, Nov 15 2021: (Start)
%C A349324 Terms cannot be a square (see the comment from Altug Alkan in A111925).
%C A349324 Terms must have at least one prime factor of the form p == 1 (mod 4), a Pythagorean prime (A002144).
%C A349324 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.
%C A349324 The special prime factor 2 has the same behavior, i.e., if the term is even, x and y must be even too. (End)
%H A349324 Jon E. Schoenfield, <a href="/A349324/b349324.txt">Table of n, a(n) for n = 1..424</a> (all terms < A346115(7) = 2474052064291275)
%H A349324 Karl-Heinz Hofmann, <a href="/A349324/a349324.gif">A 3D animation of the terms and the least one with 7 solutions</a>
%e A349324 15697403475^2 = 13780596525^2 +  86700^4
%e A349324               = 13195420077^2 +  92208^4
%e A349324               = 11468350875^2 + 103530^4
%e A349324               = 10710751443^2 + 107124^4
%e A349324               =  9221086875^2 + 112710^4
%e A349324               =  4878327597^2 + 122148^4.
%Y A349324 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).
%Y A349324 Cf. A002144 (p == 1 (mod 4)), A002145 (p == 3 (mod 4)).
%K A349324 nonn
%O A349324 1,1
%A A349324 _Jon E. Schoenfield_, Nov 14 2021