OFFSET
1,2
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
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.
EXAMPLE
(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)
PROG
(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
CROSSREFS
KEYWORD
nonn
AUTHOR
John W. Layman, Mar 03 2006
EXTENSIONS
Edited by M. F. Hasler, Jun 12 2019
STATUS
approved