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%I #29 Apr 10 2023 14:54:29
%S 7,8,15,34,71,132,223,350,519,736,1007,1338,1735,2204,2751,3382,4103,
%T 4920,5839,6866,8007,9268,10655,12174,13831,15632,17583,19690,21959,
%U 24396,27007,29798,32775,35944,39311,42882,46663,50660,54879,59326
%N a(n) = n^3 + 7.
%C These numbers cannot be perfect squares. - _Cino Hilliard_, Sep 03 2006
%C [The following short proof was supplied by _Don Reble_. - _N. J. A. Sloane_, Apr 10 2023]
%C Proof that n^3+7 <> k^2 for all integers n,k.
%C Assume y^2 - x^3 = 7 has an integer solution.
%C Modulo 4, we have {0,1,0,1} - {0,1,0,3} == 3; y is even and x is odd.
%C y^2+1 = x^3+8 = (x+2) [(x-1)^2+3]. Let z = (x-1)^2+3 == 3 mod 4.
%C The 1-mod-4 numbers are closed under multiplication, so z has a prime factor p == 3 mod 4.
%C That p divides y^2+1; y^2 == -1 mod p.
%C But (quadratic reciprocity) there is no square root of -1 modulo p.
%C That refutes the assumption.
%H Vincenzo Librandi, <a href="/A084377/b084377.txt">Table of n, a(n) for n = 0..1000</a>
%H Cino Hilliard, <a href="https://web.archive.org/web/20080621104333/http://groups.msn.com:80/BC2LCC/n37ltk2.msnw">Proof that a cube plus 7 cannot be a square</a>
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).
%F G.f.: (7 - 20*x + 25*x^2 - 6*x^3)/(1 - x)^4. - _Vincenzo Librandi_, Jun 10 2016
%F a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n>3. - _Vincenzo Librandi_, Jun 10 2016
%t Table[n^3 + 7, {n, 0, 60}] (* _Vladimir Joseph Stephan Orlovsky_, Apr 15 2011 *)
%o (PARI) a(n) = n^3 + 7;
%o (Magma) [n^3+7: n in [0..50]]; // _Vincenzo Librandi_, Jun 10 2016
%Y Cf. A000578, A084378, A084381.
%K easy,nonn
%O 0,1
%A _Cino Hilliard_, Jun 23 2003
%E More terms from _Franklin T. Adams-Watters_, Aug 29 2006
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