A047694 x such that y^2 = C(x,0) + C(x,1) + C(x,2) + C(x,3) is solvable in integers.
-1, 0, 2, 7, 15, 74, 767
Offset: 0
Examples
For x = -1, C(x, k) := x*(x-1)*...*(x-k+1)/k! = (-1)^k for all integers k >= 0, so the right hand side is zero and y = 0 is a solution. For smaller x values, the r.h.s. is negative and there can't be a solution, therefore the first term is a(1) = -1. For x = 0, C(x, k) = { 1 if k=0, else 0 }, so the r.h.s. is 1 and y = +-1 is a solution, whence a(2) = 0. For x = 1, the r.h.s. equals 1 + 1 + 0 + 0 = 2 and there is no integer solution y. For x = 2, the r.h.s. equals 1 + 2 + 1 + 0 = 4 and we have the solutions y = +-2, whence a(3) = 2.
References
- R. K. Guy, Unsolved Problems in Number Theory, Section D3.
Links
- Andrew Bremner, An equation of Mordell, Math. Comp., 29 (1975), pp. 925-928.
- W. Ljunggren, A diophantine problem, J. London Math. Soc. (2), 3 (1971), pp. 385-391.
Crossrefs
Cf. A047695 (the corresponding y values).
Programs
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Mathematica
Select[Range[-10, 10^3], IntegerQ[Sqrt[((# + 1)(#^2 - # + 6))/6]] &] (* Alonso del Arte, Sep 13 2011 *)
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PARI
A047694=[n|n<-[-1..777],issquare(binomial(n+1,3)+n+1)] \\ M. F. Hasler, Jun 22 2024
Formula
x such that 6y^2 = (x + 1)(x^2 - x + 6) has solutions in integers.
Comments