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If one calculates values of the polynomial $$f(x) := x^4\!+\!x^2\!+\!1$$ on the successive positive integers $n$ , one may observe an interesting thing when factoring the values:
$f(1) = 3$
$f(2) = 21 = 3\cdot 7$
$f(3) = 91 = 7\cdot 13$
$f(4) = 273 = 3\cdot 7\cdot 13$
$f(5) = 651 = 3\cdot 7\cdot31$
$f(6) = 1333 = 31\cdot 43$
$f(7) = 2451 = 3\cdot 19\cdot 43$
$f(8) = 4161 = 3\cdot 19\cdot 73$
$f(9) = 6643 = 7\cdot 13\cdot 73$
$f(10) = 10101 = 3\cdot 7\cdot 13\cdot 37$
$f(11) = 14763 = 3\cdot 7\cdot 19\cdot 37$
$\mbox{ . . . }$
It seems as if two consecutive values always have at least one common odd prime factor, i.e. they have the greatest common divisor $> 2$ .
It is indeed true. The reason of this fact is not particularly deep. It is easily understood if we can factorize this polynomial: $$f(x) = (x^2\!-\!x\!+\!1)(x^2\!+\!x\!+\!1)$$ Then
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(1) |
and the next value is
![$\displaystyle f(n\!+\!1) = [(n\!+\!1)^2\!-\!(n\!+\!1)\!+\!1][(n\!+\!1)^2\!+\!(n\!+\!1)\!+\!1] = (n^2\!+\!3n\!+\!3)(n^2\!+\!n\!+\!1).$ $\displaystyle f(n\!+\!1) = [(n\!+\!1)^2\!-\!(n\!+\!1)\!+\!1][(n\!+\!1)^2\!+\!(n\!+\!1)\!+\!1] = (n^2\!+\!3n\!+\!3)(n^2\!+\!n\!+\!1).$](http://images.planetmath.org:8080/cache/objects/7307/js/img2.png) |
(2) |
Thus $f(n)$ and $f(n\!+\!1)$ have as their common factor at least the number $n^2\!+\!n\!+\!1$ , which is $\geqq 3$ .
Moreover, we may show that the greatest common divisor of $f(n)$ and $f(n\!+\!1)$ is $n^2\!+\!n\!+\!1$ except in the case $n \equiv 3 \pmod{7}$ where it is $7(n^2\!+\!n\!+\!1)$ .
Let $d$ be a common divisor, greater than 1, of the first factors $n^2\!-\!n\!+\!1$ and $n^2\!+\!3n\!+\!3$ of (1) and (2). It's clear that $d$ is odd. Then $d$ must divide the difference $(n^2\!+\!3n\!+\!3)-(n^2\!-\!n\!+\!1) = 2(2n\!+\!1)$ and the sum $(n^2\!+\!3n\!+\!3)+3(n^2\!-\!n\!+\!1)= 4n^2\!+\!6$ and hence also the
difference $2(2n\!+\!1)+(4n^2\!+\!6)-(2n\!+\!1)^2 = 7$ . It means that $d = 7$ . Thus the only possible common prime factor of $n^2\!-\!n\!+\!1$ and $n^2\!+\!3n\!+\!3$ is 7. If we denote $n = 7k+r$ where $r\in\{0,\,1,\,...,\,6\}$ , we see that $$n^2\!-\!n\!+\!1 = 49k^2\!+\!14kr\!-\!7k\!+\!r^2\!-\!r\!+\!1 \,\,\overline{\equiv}\,\, r^2\!-\!r\!+\!1 \pmod{7},$$ $$n^2\!+\!3n\!+\!3 = 49k^2\!+\!14kr\!+\!21k\!+\!r^2\!+\!3r\!+\!3 \,\,\overline{\equiv}\,\, r^2\!+\!3r\!+\!3 \pmod{7}.$$ It's easy to check that the right hand sides of these polynomial congruences are divisible by 7 only if $r = 3$ , i.e. if $n \equiv 3 \pmod{7}$ .
Note. The sequence formed by the successive values of $\gcd(f(n), f(n\!+\!1))$ is number A111002 in Sloane's register (http://www.research.att.com/~njas/sequences/).
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