A115169 Integers b > 0 for which there exists a positive integer a <= b such that (a^2 + b^2)/(1 + ab) is an integer.

1, 8, 27, 30, 64, 112, 125, 216, 240, 343, 418, 512, 729, 1000, 1020, 1331, 1560, 1728, 2133, 2197, 2744, 3120, 3375, 4096, 4913, 5822, 5832, 6859, 7770, 8000, 9261, 10648, 12167, 13824, 15625, 16256, 16800, 17576, 18957, 19683

Offset: 1

Author

John W. Layman, Mar 03 2006

Keywords

• nonn

Comments

All positive cubes are in this sequence.

Indeed, if b = k^3, then for a = k, we have a^2 + b^2 = k^2 + k^6 = (1 + k^4)*k^2 = (a*b + 1)*a^2. More generally, if the ratio (a^2 + b^2)/(a*b + 1) is an integer, it is equal to gcd(a,b)^2, thus in particular a perfect square. (This was Question 6 in the 1988 IMO.) All solutions (a,b) are member of a sequence {(x(n), x(n+1)); n = 1,2,...} where x = (0, k, k^3, k^5 - k, ...) with x(n+1) = k^2*x(n) - x(n-1) and some k >= 2, cf. A052530 for k = 2, A065100 for k = 3. (One might consider >= 0 instead > 0 in the definition, but a = 0 yields a solution for any b.) - M. F. Hasler, Jun 12 2019

Examples

			(2^2+8^2)/(1+2*8) = 68/17 = 4, an integer, so 8 is a term of the series.
From _M. F. Hasler_, Jun 12 2019: (Start)
The list of solutions starts:
     a      b     a^2+b^2   a*b+1   ratio
   ----------------------------------------
     1      1          2       2       1
     8      2         68      17       4
    27      3        738      82       9
    30      8        964     241       4
    64      4       4112     257      16
   112     30      13444    3361       4
   125      5      15650     626      25
   216      6      46692    1297      36
   240     27      58329    6481       9
   343      7     117698    2402      49
   418    112     187268   46817       4
   512      8     262208    4097      64
   729      9     531522    6562      81
  1000     10    1000100   10001     100
  1020     64    1044496   65281      16
(End)
		

Links

• Charles R Greathouse IV, Table of n, a(n) for n = 1..500
• Steve Chow, You, Me and The Legend of Question Six, BlackPenRedPen, YouTube, Apr 20 2020.
• I. Lauko, G. Pinter and L. Pinter, Another Step Further... On a Problem of the 1988 IMO, Math. Mag. 79 (2006), 45-53.
• Simon Pampena, The Legend of Question Six, Numberphile, YouTube, Aug 16 2016.

Crossrefs

Cf. A000578 (cubes), A052530 (subsequence of terms for ratio 2^2, for n >= 2), A065100 (subsequence of terms for ratio 3^2).

Programs

PARI

isok(n) = for(m=0, n, if (denominator((m^2+n^2)/(1+m*n))==1, return(1))); return (0); \\ Michel Marcus, Sep 18 2017

PARI

is_A115169(n)=for(a=1,n\3+1,(a^2+n^2)%(1+a*n)||return(1)) \\ M. F. Hasler, Jun 12 2019

PARI

is(n)=my(s=sqrtnint(n,3),n2=n^2); for(b=1,s, if((n2+b^2)%(n*b+1)==0, return(1))); for(K=2,sqrtint((n2+(s+1)^2)\(n*s+n+1)), my(k=K^2); if(issquare(k^2*n2-4*n2+4*k), return(1))); 0 \\ Charles R Greathouse IV, Nov 08 2021

Extensions

Edited by M. F. Hasler, Jun 12 2019