Quickie #8

**Soroban** · January 6th 2007, 11:10 AM

Prove that for any positive integer $n$

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

**ThePerfectHacker** · January 6th 2007, 02:20 PM

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!!!

**galactus** · January 6th 2007, 04:39 PM

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.

**Soroban** · January 6th 2007, 07:35 PM

Hello, Hacker and Galactus!

I like both your solutions.

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.

Thread: Quickie #8

LinkBack

Thread Tools

Display

Quickie #8

Similar Math Help Forum Discussions

Quickie #15

Quickie #14

Quickie #13

Quickie #12

Quickie #1

Search Tags