# Quickie #8

• Jan 6th 2007, 11:10 AM
Soroban
Quickie #8
Prove that for any positive integer $\displaystyle n$

. . $\displaystyle n^4 + 2n^3 + 2n^2 + 2n + 1$ is never a perfect square.

• Jan 6th 2007, 02:20 PM
ThePerfectHacker
Quote:

Originally Posted by Soroban
Prove that for any positive integer $\displaystyle n$

. . $\displaystyle n^4 + 2n^3 + 2n^2 + 2n + 1$ is never a perfect square.

$\displaystyle (n^4+2n^2+1)+(2n^3+2n)$
$\displaystyle (n^2+1)^2+2n(n^2+1)$
$\displaystyle (n^2+1)((n^2+1)+2n)$
$\displaystyle (n^2+1)(n^2+2n+1)$
$\displaystyle (n^2+1)(n+1)^2$
Assume it is a square....
Theorem: If a product of a positive integer and a square is is square then that positive integer itself must be a square.

Thus,
$\displaystyle n^2+1=m^2$
$\displaystyle m^2-n^2=1$
$\displaystyle (m+n)(m-n)=1$
Trivial-Improper factorization,
$\displaystyle m+n=1$
$\displaystyle m-n=1$
Thus,
$\displaystyle n=0$.
But the problem says positive integer.... a contradiction.

This is mine 4:):):)th Post!!!
• Jan 6th 2007, 04:39 PM
galactus
Here's a thought:

If you factor:

$\displaystyle (n+1)^{2}(n^{2}+1)=k^{2}$

Take the square root:

$\displaystyle (n+1)\sqrt{n^{2}+1}=k$

If n is a positive integer, then k can not be an integer, therefore, can not be a perfect square.
• Jan 6th 2007, 07:35 PM
Soroban
Hello, Hacker and Galactus!

The "Quickie" solution is surprisingly clever
. . but almost impossible to derive.

Let $\displaystyle N \:=\:n^4 + 2n^3 + 2n^2 + 2n + 1$

Since $\displaystyle n^4 + 2n^3 + n^2 \:< \:n^4 + 2n^3 + 2n^2 + 2n + 1$
. . we have: .$\displaystyle (n^2 + n)^2 \:< \:N$

Since $\displaystyle n^4 + 2n^3 + 2n^2 + 2n + 1 \:< \:n^4 + 2n^3 + 3n^2 + 2n + 1$
. . we have: .$\displaystyle N \:< \:(n^2 + n + 1)^2$

Hence: .$\displaystyle (n^2 + n)^2 \:< \:N \:< \:(n^2 + n + 1)^2$
. . and $\displaystyle N$ lies between two consecutive squares.