A257874 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.

3, 6, 8, 10, 12, 15, 17, 21, 24, 27, 31, 33, 38, 41, 43, 48, 50, 52, 54, 57, 59, 62, 66, 69, 71, 73, 75, 78, 80, 82, 85, 90, 93, 95, 97, 99, 101, 103, 105, 111, 115, 117, 119, 124, 127, 131, 133, 136, 138, 141, 143, 145, 147, 150, 153, 155, 157, 162, 164, 168

Offset: 1

Author

Michel Lagneau, May 11 2015

Keywords

• nonn

Comments

The sequence is infinite. Proof:

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

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.

Keeping in mind that n^2-n+1 is odd, we have

gcd(n^2+n+1,n^2-n+1) = gcd(2n,n^2-n+1)= gcd(n,n^2-n+1)= 1.

Therefore q(n) is different from q(n-1).

To prove the result, it suffices to show that the set

S = {n in Z | n>=2 and q(n) > q(n-1) and q(n) > q(n+1)}

is infinite, since for each n in S one has

p(n) = max{q(n),q(n-1)} = q(n) = max{q(n),q(n+1)} = p(n+1).

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.

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)= k+1 such that q(l)>q(l+1). By the minimality of l, we have q(l-1)= k + 1 > k >= m, this contradicts the maximality of m, and hence S is indeed infinite.

Examples

			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.
		

Links

• Charles R Greathouse IV, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A002061, A059826.

Programs

Maple

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:

Mathematica

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 *)
SequencePosition[Table[FactorInteger[n^4+n^2+1][[-1,1]],{n,200}],{x_,x_}][[;;,1]] (* Harvey P. Dale, Nov 06 2024 *)

PARI

gpf(n)=if(n<4, return(n)); my(f=factor(n)[,1]); f[#f]
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