# Perfect square

• Sep 13th 2009, 07:50 AM
dhiab
Perfect square
Find all naturel numbers n such as : $n^4 - 4n^3 + 22n^2 - 36n + 18$ is the Perfect square .
• Sep 13th 2009, 08:34 AM
Soroban
Hello, dhiab!

Is there a typo?
If the last constant is 81, there is a simple solution.

Quote:

Find all natural numbers $n$ such that: . $n^4 - 4n^3 + 22n^2 - 36n + {\color{red}81}$ . is a perfect square.

Let: . $(n^2 + an + b)^2 \:=\:n^4 - 4n^3 + 22n^2 - 36n + 81$

Expand: . $n^4 + 2an^3 + (a^2+2b)n^2 + 2abn + b^2 \;=\;n^4 - 4n^3 + 22n^2 - 36n + 81$

And we have: . $\begin{array}{c} 2a \:=\:-4 \\ a^2+2b \:=\:22 \\ 2ab \:=\:-36 \\ b^2 \:=\:81 \end{array} \quad\Rightarrow \quad a \:=\:-2,\;b \:=\:9$

Therefore: . $(n^2 - 2n + 9)^2 \;=\;n^4 - 4n^3 + 22n^2 - 37n + 81$

The expression is a square for all natural numbers $n.$