arithmetic

tnhh · New Member Offline Joined: 28 Oct 2009 Posts: 4

Suppose that d is a factor of $n^4 + 2n^2 + 2$ such that $d>n^2 + 1$ ,where n is some natural number n>1. Prove that $d > n^2 + 1 + \sqrt {n^2 + 1}$

**Posted:** Wed Oct 28, 2009 7:17 am

cnyd

$n^{2}+1=x$ $\implies$ $d|x^{2}+1$ and $d>x$
$\implies$ $d=x+a$
$\frac{x^{2}+1}{x+a}=x-a+\frac{a^{2}+1}{x+a}$ $\implies$ $x+a|a^{2}+1$
$a^{2}-a+1>x$ $\implies$ $a>\sqrt{x}$ QED Wink

**Posted:** Wed Oct 28, 2009 11:35 am

tnhh · New Member Offline Joined: 28 Oct 2009 Posts: 4

Why do you have this?
$x + a|a^{2} + 1$
$=> a^{2} - a + 1 > x$

**Posted:** Wed Nov 04, 2009 7:08 am

Yongyi781

$x + a|a^2 + 1\implies a^2 + 1 \geq x + a$ , or $a^2 - a + 1 \geq x$ . But $x=n^2+1$ , so $a^2-a+1\geq n^2+1\implies a^2-a\geq n^2$ . Clearly equality cannot occur so we have $a^2-a>n^2$ , and $a^2-a+1>x$ .

**Posted:** Thu Nov 05, 2009 4:35 am

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