A175155 Numbers m satisfying m^2 + 1 = x^2 * y^3 for positive integers x and y.

0, 682, 1268860318, 1459639851109444, 2360712083917682, 86149711981264908618, 4392100110703410665318, 8171493471761113423918890682, 15203047261220215902863544865414318, 5484296027914919579181500526692857773246, 28285239023397517753374058381589688919682, 12439333951782387734360136352377558500557329868

Offset: 1

Author

Michel Lagneau, Feb 27 2010

Keywords

• nonn

Comments

This sequence is infinite. The fundamental solution of m^2 + 1 = x^2 y^3 is (m,x,y) = (682,61,5), which means the Pellian equation m^2 - 125x^2 = -1 has the solution (m,x) = (682,61) = (m(1),x(1)). This Pellian equation admits an infinity of solutions (m(2k+1),x(2k+1)), k=1,2,..., given by the following recursive relation, starting with m(1)=682, x(1)= 61: m(2k+1) + x(2k+1)*sqrt(125) = (m(1) + x(1)*sqrt(125))^(2k+1).

Squares of these terms are in A060355, since both a(n)^2 and a(n)^2 + 1 are powerful (A001694). - Charles R Greathouse IV, Nov 16 2012

It appears that y = A077426. - Robert G. Wilson v, Nov 16 2012

Also m^2 + 1 is powerful. Other solutions arise from solutions x to x^2 - k^3*y^2 = -1. - Georgi Guninski, Nov 17 2012

Although it is believed that the b-file is complete for all terms m < 10^100, the search only looked for y < 100000. - Robert G. Wilson v, Nov 17 2012

Examples

			For m=682, m^2 + 1 = 465125 = 61^2 * 5^3.
		

References

• Albert H. Beiler, "The Pellian" (Chap. 22), Recreations in the Theory of Numbers, 2nd ed. NY: Dover, 1966.
• A. Cayley, Report of a committee appointed for the purpose of carrying on the tables connected with the Pellian equation ..., Collected Mathematical Papers. Vols. 1-13, Cambridge Univ. Press, London, 1889-1897, Vol. 13, pp. 430-443.
• J. M. De Koninck, Ces nombres qui nous fascinent, Ellipses, 2008, p. 108.

Links

• Robert G. Wilson v, Table of n, a(n) for n = 1..292. [This list has not been proved to be complete! - _N. J. A. Sloane_, Nov 17 2012]
• E. E. Whitford, The Pell equation, New York, 1912.
• Wikipedia, Pell's equation

Crossrefs

Pellian equation: A006704, A006702, A006705, A006703, A003153, A001571, A001570.

Programs

Maple

C:=array(0..20,0..20):C[1,1]=1: C[2,1]=1: n1:=682:x1:=61:for nn from 1 by 2 to 15 do:s:=0:for i from 2 to 15 do:for j from 1 to i do:C[i,j]:= C[i-1,j] + C[i-1,j-1]: od:od:for n from 1 by 2 to nn+1 do:s:=s + C[nn+1,n] * n1^(nn-n+1)*x1^(n-1)*125^((n-1)/2):od:print (s):od: # Michel Lagneau
# 2nd program R. J. Mathar, Mar 16 2016:
# print (nonsorted!) all solutions of A175155 up to search limit
with(numtheory):
# upper limit for solutions n
nsearchlim := 10^40 :
A175155y := proc(y::integer)
    local disc;
    disc := y^3 ;
    cfrac(sqrt(disc),periodic,quotients) ;
end proc:
for y from 2 do
    if issqrfree(y) then
        # find continued fraction for x^2-(y^3=disc)*y^2=-1, sqrt(disc)
        cf := A175155y(y) ;
        nlen :=  nops(op(2,cf)) ;
        if type(nlen,odd) then
            # fundamental solution
            fuso := numtheory[nthconver](cf,nlen-1) ;
            fusolx := numer(fuso) ;
            fusoly := denom(fuso) ;
            solx := fusolx ;
            soly := fusoly ;
            while solx <= nsearchlim do
                rhhs := solx^2-y^3*soly^2 ;
                if rhhs = -1 then
                    # print("n=",solx,"x=",soly,"y=",y^3) ;
                    print(solx) ;
                end if;
                # solutions from fundamental solution
                tempx := fusolx*solx+y^3*fusoly*soly ;
                tempy := fusolx*soly+fusoly*solx ;
                solx := tempx ;
                soly := tempy ;
            end do;
        end if;
    fi;
end do:

Mathematica

nmax = 10^50; ymax = 100; instances = 10; fi[y_] := n /. FindInstance[0 <= n <= nmax && x > 0 && n^2 + 1 == x^2*y^3, {n, x}, Integers, instances]; yy = Select[Range[1, ymax, 2], !IntegerQ[Sqrt[#]] && OddQ[ Length[ ContinuedFraction[Sqrt[#]][[2]]]]&]; Join[{0}, fi /@ yy // Flatten // Union // Most] (* Jean-François Alcover, Jul 12 2017 *)

PARI

is(n)=ispowerful(n^2+1) \\ Charles R Greathouse IV, Nov 16 2012

Formula

m(1)=682, x(1) = 61 and m(2k+1) + x(2k+1)*sqrt(125) = (m(1) + x(1)*sqrt(125))^(2k+1) m(2k+1) = C(2k+1,0) * m(1)^(2k+1) + C(2k+1,2)*m(1)^(2k-1)*x(1)^2 + ...

Extensions

Added condition that x and y must be positive. Added missing initial term 0. Added warning that b-file has not been proved to be correct - there could be missing entries. - N. J. A. Sloane, Nov 17 2012