| Version 8 |
Version 7 |
| If one calculates values of the polynomial |
If one calculates values of the polynomial |
| $$f(x) := x^4\!+\!x^2\!+\!1$$ |
$$f(x) := x^4\!+\!x^2\!+\!1$$ |
| on the successive positive integers $n$, one may observe an interesting thing when factoring the values: |
on the successive positive integers $n$, one may observe an interesting thing when factoring the values: |
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| $f(1) = 3$\\ |
$f(1) = 3$\\ |
| $f(2) = 21 = 3\cdot 7$\\ |
$f(2) = 21 = 3\cdot 7$\\ |
| $f(3) = 91 = 7\cdot 13$\\ |
$f(3) = 91 = 7\cdot 13$\\ |
| $f(4) = 273 = 3\cdot 7\cdot 13$\\ |
$f(4) = 273 = 3\cdot 7\cdot 13$\\ |
| $f(5) = 651 = 3\cdot\ 7\cdot31$\\ |
$f(5) = 651 = 3\cdot\ 7\cdot31$\\ |
| $f(6) = 1333 = 31\cdot 43$\\ |
$f(6) = 1333 = 31\cdot 43$\\ |
| $f(7) = 2451 = 3\cdot 19\cdot 43$\\ |
$f(7) = 2451 = 3\cdot 19\cdot 43$\\ |
| $f(8) = 4161 = 3\cdot 19\cdot 73$\\ |
$f(8) = 4161 = 3\cdot 19\cdot 73$\\ |
| $f(9) = 6643 = 7\cdot 13\cdot 73$\\ |
$f(9) = 6643 = 7\cdot 13\cdot 73$\\ |
| $f(10) = 10101 = 3\cdot 7\cdot 13\cdot 37$\\ |
$f(10) = 10101 = 3\cdot 7\cdot 13\cdot 37$\\ |
| $f(11) = 14763 = 3\cdot 7\cdot 19\cdot 37$\\ |
$f(11) = 14763 = 3\cdot 7\cdot 19\cdot 37$\\ |
| $\mbox{ . . . }$ |
$\mbox{ . . . }$ |
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| 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 seems as if two consecutive values always have at least one common odd prime factor, i.e. they have the greatest common divisor $> 2$. |
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| It is indeed true.\, The reason of this fact is not particularly deep.\, It is easily understood if we can \PMlinkname{factorize this polynomial}{GroupingMethodForFactorizingPolynomials}: |
It is indeed true.\, The reason of this fact is not particularly deep.\, It is easily understood if we can \PMlinkname{factorize this polynomial}{GroupingMethodForFactorizingPolynomials}: |
| $$f(x) = (x^2\!-\!x\!+\!1)(x^2\!+\!x\!+\!1)$$ |
$$f(x) = (x^2\!-\!x\!+\!1)(x^2\!+\!x\!+\!1)$$ |
| Then |
Then |
| \begin{align} |
\begin{align} |
| f(n) = (n^2\!-\!n\!+\!1)(n^2\!+\!n\!+\!1), |
f(n) = (n^2\!-\!n\!+\!1)(n^2\!+\!n\!+\!1), |
| \end{align} |
\end{align} |
| and the next value is |
and the next value is |
| \begin{align} |
\begin{align} |
| f(n\!+\!1) = [(n\!+\!1)^2\!-\!(n\!+\!1)\!+\!1][(n\!+\!1)^2\!+\!(n\!+\!1)\!+\!1] = (n^2\!+\!3n\!+\!3)(n^2\!+\!n\!+\!1). |
f(n\!+\!1) = [(n\!+\!1)^2\!-\!(n\!+\!1)\!+\!1][(n\!+\!1)^2\!+\!(n\!+\!1)\!+\!1] = (n^2\!+\!3n\!+\!3)(n^2\!+\!n\!+\!1). |
| \end{align} |
\end{align} |
| Thus $f(n)$ and $f(n\!+\!1)$ have as their common factor at least the number $n^2\!+\!n\!+\!1$, which is $\geqq 3$. |
Thus $f(n)$ and $f(n\!+\!1)$ have as their common factor at least the number $n^2\!+\!n\!+\!1$, which is $\geqq 3$. |
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| Moreover, we may show that the greatest common factor of $f(n)$ and $f(n\!+\!1)$ is $n^2\!+\!n\!+\!1$ except in the case\, $n \equiv 3 \pmod{7}$\, where it is $7(n^2\!+\!n\!+\!1)$. |
Moreover, we may show that the greatest common factor of $f(n)$ and $f(n\!+\!1)$ is $n^2\!+\!n\!+\!1$ except in the case\, $n \equiv 3 \pmod{7}$\, where it is $7(n^2\!+\!n\!+\!1)$. |
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| 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 \PMlinkname{odd}{OddNumber}.\, 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 |
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 \PMlinkname{odd}{OddNumber}.\, 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 |
| $$n^2\!-\!n\!+\!1 = 49k^2\!+\!7k\!+\!r^2\!-\!r\!+\!1 \,\,\overline{\equiv}\,\, r^2\!-\!r\!+\!1 \pmod{7},$$ |
$$n^2\!-\!n\!+\!1 = 49k^2\!+\!7k\!+\!r^2\!-\!r\!+\!1 \,\,\overline{\equiv}\,\, r^2\!-\!r\!+\!1 \pmod{7},$$ |
| $$n^2\!+\!3n\!+\!3 = 49k^2\!+\!14kr\!+\!21k\!+\!r^2\!+\!3r\!+\!3 \,\,\overline{\equiv}\,\, r^2\!+\!3r\!+\!3 \pmod{7}.$$ |
$$n^2\!+\!3n\!+\!3 = 49k^2\!+\!14kr\!+\!21k\!+\!r^2\!+\!3r\!+\!3 \,\,\overline{\equiv}\,\, r^2\!+\!3r\!+\!3 \pmod{7}.$$ |
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It's easy to check that the right hand \PMlinkescapetext{sides} of these polynomial congruences are divisible by 7 only if\, $r = 3$,\, i.e. if\, $n \equiv 3 \pmod{7}$.
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It's easy to check that the right hand \PMlinkescapetext{side} of these polynomial congruences are divisible by 7 only if\, $r = 3$,\, i.e. if\, $n \equiv 3 \pmod{7}$.
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