# Quickie #8

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

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

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

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

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

$(n^4+2n^2+1)+(2n^3+2n)$
$(n^2+1)^2+2n(n^2+1)$
$(n^2+1)((n^2+1)+2n)$
$(n^2+1)(n^2+2n+1)$
$(n^2+1)(n+1)^2$
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,
$n^2+1=m^2$
$m^2-n^2=1$
$(m+n)(m-n)=1$
Trivial-Improper factorization,
$m+n=1$
$m-n=1$
Thus,
$n=0$.
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:

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

Take the square root:

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

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

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

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

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