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%I #28 Nov 09 2021 13:38:25
%S 1,8,27,30,64,112,125,216,240,343,418,512,729,1000,1020,1331,1560,
%T 1728,2133,2197,2744,3120,3375,4096,4913,5822,5832,6859,7770,8000,
%U 9261,10648,12167,13824,15625,16256,16800,17576,18957,19683
%N Integers b > 0 for which there exists a positive integer a <= b such that (a^2 + b^2)/(1 + ab) is an integer.
%C All positive cubes are in this sequence.
%C 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
%H Charles R Greathouse IV, <a href="/A115169/b115169.txt">Table of n, a(n) for n = 1..500</a>
%H Steve Chow, <a href="https://youtu.be/usEQRx4J_ew">You, Me and The Legend of Question Six</a>, BlackPenRedPen, YouTube, Apr 20 2020.
%H I. Lauko, G. Pinter and L. Pinter, <a href="http://www.jstor.org/stable/27642901">Another Step Further... On a Problem of the 1988 IMO</a>, Math. Mag. 79 (2006), 45-53.
%H Simon Pampena, <a href="https://www.youtube.com/watch?v=Y30VF3cSIYQ">The Legend of Question Six</a>, Numberphile, YouTube, Aug 16 2016.
%e (2^2+8^2)/(1+2*8) = 68/17 = 4, an integer, so 8 is a term of the series.
%e From _M. F. Hasler_, Jun 12 2019: (Start)
%e The list of solutions starts:
%e a b a^2+b^2 a*b+1 ratio
%e ----------------------------------------
%e 1 1 2 2 1
%e 8 2 68 17 4
%e 27 3 738 82 9
%e 30 8 964 241 4
%e 64 4 4112 257 16
%e 112 30 13444 3361 4
%e 125 5 15650 626 25
%e 216 6 46692 1297 36
%e 240 27 58329 6481 9
%e 343 7 117698 2402 49
%e 418 112 187268 46817 4
%e 512 8 262208 4097 64
%e 729 9 531522 6562 81
%e 1000 10 1000100 10001 100
%e 1020 64 1044496 65281 16
%e (End)
%o (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
%o (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
%o (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
%Y Cf. A000578 (cubes), A052530 (subsequence of terms for ratio 2^2, for n >= 2), A065100 (subsequence of terms for ratio 3^2).
%K nonn
%O 1,2
%A _John W. Layman_, Mar 03 2006
%E Edited by _M. F. Hasler_, Jun 12 2019
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