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A340129 a(n) is the number of solutions of the Diophantine equation x^2 + y^2 = z^5 + z, gcd(x, y, z) = 1, x <= y, where z = A008784(n).

%I A340129 #55 Jan 22 2021 07:34:44
%S A340129 1,1,2,4,2,2,4,4,2,2,4,4,4,4,4,2,4,2,2,2,16,4,4,4,2,4,2,4,4,8,8,8,4,4,
%T A340129 4,2,8,4,16,4,16,4,2,4,16,4,4,16,4,8,8,8,4,4,8,4,8,4,4,4,16,4,4,8,2,
%U A340129 16,2,32,2,16,4,4,2,4,8,16,4,8,4,8,4,4,8,4,16
%N A340129 a(n) is the number of solutions of the Diophantine equation x^2 + y^2 = z^5 + z, gcd(x, y, z) = 1, x <= y, where z = A008784(n).
%C A340129 The idea of this sequence comes from the 6th problem of the 21st British Mathematical Olympiad in 1985 where it is asked to show that this equation has infinitely many solutions (see link Olympiads and reference Gardiner).
%C A340129 Indeed, this Diophantine equation x^2 + y^2 = z^5 + z with gcd(x, y, z) = 1 has solutions iff z is in A008784.
%C A340129 When z is in A008784, there exist (u, v), gcd(u, v) = 1 such that z = u^2 + v^2; then, (u*z^2-v)^2 + (u+v*z^2)^2 = z^5 + z. Hence, with x = min(u*z^2-v, u+v*z^2) and y = max(u*z^2-v, u+v*z^2), the equation x^2 + y^2 = z^5 + z is satisfied. So this equation has infinitely many solutions since it has at least one solution for each term of A008784.
%C A340129 For instance, for z = 10 we have:
%C A340129   with (u,v) = (1,3), then x = 1*10^2-3 = 97 and y = 1+3*10^2 = 301,
%C A340129   with (u,v) = (3,1), then x = 3+1*10^2 = 103 and y = 3*10^2-1 = 299,
%C A340129   so finally 97^2 + 301^2 = 103^2 + 299^2 = 10^5 + 10.
%C A340129 Note that some z, among them 10, have other solutions not of this form.
%D A340129 A. Gardiner, The Mathematical Olympiad Handbook: An Introduction to Problem Solving, Oxford University Press, 1997, reprinted 2011, Problem 6, pp. 63 and 167-168 (1985).
%H A340129 Amiram Eldar, <a href="/A340129/b340129.txt">Table of n, a(n) for n = 1..10000</a>
%H A340129 British Mathematical Olympiad, <a href="https://bmos.ukmt.org.uk/home/bmo-1985.pdf">1985 - Problem 6</a>.
%H A340129 <a href="/index/O#Olympiads">Index to sequences related to Olympiads</a>.
%e A340129 For z = A008784(1) = 1, 1^2 + 1^2 = 1^5 + 1 is the only solution, so a(1) = 1.
%e A340129 For z = A008784(3) = 5, 23^2 + 51^2 = 27^2 + 49^2 = 5^5 + 5 so a(3) = 2.
%e A340129 For z = A008784(4) = 10, (97, 301, 10), (103, 299, 10), (119, 293, 10) and (163, 271, 10) are solutions, so a(4) = 4.
%t A340129 f[n_] := Length @ Solve[x^2 + y^2 == n^5 + n && GCD @@ {x, y, n} == 1 && 0 <= x <= y, {x, y}, Integers]; f /@ Select[Range[500], IntegerExponent[#, 2] < 2 && AllTrue[FactorInteger[#][[;; , 1]], Mod[#1, 4] < 3 &] &] (* _Amiram Eldar_, Jan 22 2021 *)
%o A340129 (PARI) f(z) = {if (issquare(Mod(-1, z)), my(nb = 0, s = z^5+z, d, j); for (i=1, sqrtint(s), if (issquare(d = s - i^2), j = sqrtint(d); if ((j<=i) && gcd([i, j, z]) == 1, nb++););); nb;);}
%o A340129 lista(nn) = {for (n=1, nn, if (issquare(Mod(-1, n)), print1(f(n), ", ")););} \\ _Michel Marcus_, Jan 20 2021
%Y A340129 Cf. A008784, A048147, A131471.
%K A340129 nonn
%O A340129 1,3
%A A340129 _Bernard Schott_, Jan 17 2021
%E A340129 More terms from _Michel Marcus_, Jan 20 2021