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).

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, 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, 16, 2, 32, 2, 16, 4, 4, 2, 4, 8, 16, 4, 8, 4, 8, 4, 4, 8, 4, 16

Offset: 1

Author

Bernard Schott, Jan 17 2021

Keywords

• nonn

Comments

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).

Indeed, this Diophantine equation x^2 + y^2 = z^5 + z with gcd(x, y, z) = 1 has solutions iff z is in A008784.

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.

For instance, for z = 10 we have:

with (u,v) = (1,3), then x = 1*10^2-3 = 97 and y = 1+3*10^2 = 301,

with (u,v) = (3,1), then x = 3+1*10^2 = 103 and y = 3*10^2-1 = 299,

so finally 97^2 + 301^2 = 103^2 + 299^2 = 10^5 + 10.

Note that some z, among them 10, have other solutions not of this form.

Examples

			For z = A008784(1) = 1, 1^2 + 1^2 = 1^5 + 1 is the only solution, so a(1) = 1.
For z = A008784(3) = 5, 23^2 + 51^2 = 27^2 + 49^2 = 5^5 + 5 so a(3) = 2.
For z = A008784(4) = 10, (97, 301, 10), (103, 299, 10), (119, 293, 10) and (163, 271, 10) are solutions, so a(4) = 4.
		

References

• 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).

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000
• British Mathematical Olympiad, 1985 - Problem 6.
• Index to sequences related to Olympiads.

Crossrefs

Cf. A008784, A048147, A131471.

Programs

Mathematica

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 *)

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;);}
lista(nn) = {for (n=1, nn, if (issquare(Mod(-1, n)), print1(f(n), ", ")););} \\ Michel Marcus, Jan 20 2021

Extensions

More terms from Michel Marcus, Jan 20 2021