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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). 2
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,3

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.

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.

EXAMPLE

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.

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 *)

PROG

(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

CROSSREFS

Cf. A008784, A048147, A131471.

Sequence in context: A021809 A210210 A117007 * A213433 A294354 A144049

Adjacent sequences:  A340126 A340127 A340128 * A340130 A340131 A340132

KEYWORD

nonn

AUTHOR

Bernard Schott, Jan 17 2021

EXTENSIONS

More terms from Michel Marcus, Jan 20 2021

STATUS

approved

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Last modified October 23 00:39 EDT 2021. Contains 348211 sequences. (Running on oeis4.)