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%I #19 Dec 11 2015 21:55:48
%S 3,6,8,10,12,15,17,21,24,27,31,33,38,41,43,48,50,52,54,57,59,62,66,69,
%T 71,73,75,78,80,82,85,90,93,95,97,99,101,103,105,111,115,117,119,124,
%U 127,131,133,136,138,141,143,145,147,150,153,155,157,162,164,168
%N Numbers n such that the largest prime divisor of n^4 + n^2 + 1 is equal to the largest prime divisor of (n+1)^4 + (n+1)^2 + 1.
%C The sequence is infinite. Proof:
%C Let p(n) denote the largest prime divisor of n^4 + n^2 + 1 and let q(n) denote the largest prime divisor of n^2 + n + 1. Then p(n) = q(n^2), and from
%C n^4 + n^2 + 1 = (n^2+1)^2 - n^2 = (n^2-n+1)(n^2+n+1)= ((n-1)^2 + (n-1)+1)(n^2+n+1) it follows that p(n) = max{q(n),q(n-1)} for n>=2.
%C Keeping in mind that n^2-n+1 is odd, we have
%C gcd(n^2+n+1,n^2-n+1) = gcd(2n,n^2-n+1)= gcd(n,n^2-n+1)= 1.
%C Therefore q(n) is different from q(n-1).
%C To prove the result, it suffices to show that the set
%C S = {n in Z | n>=2 and q(n) > q(n-1) and q(n) > q(n+1)}
%C is infinite, since for each n in S one has
%C p(n) = max{q(n),q(n-1)} = q(n) = max{q(n),q(n+1)} = p(n+1).
%C Suppose on the contrary that S is finite. Since q(2) = 7 < 13 = q(3) and q(3) = 13 > 7 = q(4), the set S is nonempty. Since it is finite, we can consider its largest element, say m.
%C Note that it is impossible that q(m)>q(m+1)>q(m+2)>... because all these numbers are positive integers, so there exists a number k>=m such that q(k)<q(k+1) (recall that q(k) is different from q(k+1)). Next observe that it is impossible to have q(k)<q(k+1)<q(k+2)<... because q((k+1)^2) = p(k+1) = max{q(k),q(k+1)}, so let us take the smallest l >= k+1 such that q(l)>q(l+1). By the minimality of l, we have q(l-1)<q(l), so l is in S. Since l >= k + 1 > k >= m, this contradicts the maximality of m, and hence S is indeed infinite.
%H Charles R Greathouse IV, <a href="/A257874/b257874.txt">Table of n, a(n) for n = 1..10000</a>
%e 3 is in the sequence because 3^4+3^2+1 = 91 = 13*7 and 4^4+4^2+1 = 273 = 13*7*3, so 13 is the greatest prime factor of both expressions.
%p with(numtheory):for n from 1 to 400 do:x1:=n^4 + n^2 + 1:x2:=(n+1)^4 + (n+1)^2+1:y1:=factorset(x1):n1:=nops(y1):y2:=factorset(x2):n2:=nops(y2):if y1[n1]=y2[n2] then printf(`%d, `,n):else fi:od:
%t fQ[n_]:=Last[FactorInteger[n^4+n^2+1]][[1]]==Last[FactorInteger[(n+1)^4+(n+1)^2+1]][[1]];Select[Range[168],fQ[#]&] (* _Ivan N. Ianakiev_, Jun 11 2015 *)
%o (PARI) gpf(n)=if(n<4, return(n)); my(f=factor(n)[,1]); f[#f]
%o is(n)=my(a=n^4+n^2+1, b=(n+1)^4 +(n+1)^2+1, g=gcd(a,b), p=gpf(g)); g>1 && p>=gpf(a/g) && p>=gpf(b/g) \\ _Charles R Greathouse IV_, May 11 2015
%Y Cf. A002061, A059826.
%K nonn
%O 1,1
%A _Michel Lagneau_, May 11 2015
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