# Thread: Write as whole square

1. ## Write as whole square

$n(n+1)(n+2)(n+3)+1$
as whole square

like writing ${x}^2+4x+4=(x+2)(x+2) = {(x+2)}^{2}$

My Try:
$n(n+1)(n+2)(n+3)+1={n}^{4}+{6n}^{3}+{11n}^{2}+6n+1$
and this equation has 2 pairs of equal roots, and they are not integers

2. Originally Posted by razemsoft21
$n(n+1)(n+2)(n+3)+1$
as whole square

like writing ${x}^2+4x+4=(x+2)(x+2) = {(x+2)}^{2}$

My Try:
$n(n+1)(n+2)(n+3)+1={n}^{4}+{6n}^{3}+{11n}^{2}+6n+1$
and this equation has 2 pairs of equal roots, and they are not integers
factorise x&#40;x&#43;1&#41;&#40;x&#43;2&#41;&#40;x&#43;3&#4 1; &#43; 1 - Wolfram|Alpha

can you please show me the way you solve it ?

4. Originally Posted by razemsoft21
can you please show me the way you solve it ?
Can you reverse engineer a solution from the answer ...? See how you go.

5. Hello, razemsoft21!

Write as a square: . $n(n+1)(n+2)(n+3)+1$

We have: . $\bigg[n(n+3)\bigg]\,\bigg[(n+1)(n+2)\bigg] + 1$

. . $=\;\bigg[n^2 + 3n\bigg]\,\bigg[n^2+3n + 2\bigg] +1$

. . $=\; \bigg[n^2 + 3n +1 - 1\bigg]\,\bigg[n^2 + 3n + 1 + 1\bigg] + 1$

. . $=\; \bigg[(n^2 + 3n +1) - 1\bigg]\,\bigg[(n^2 + 3n + 1) + 1\bigg] + 1$

. . $=\;\bigg[(n^2+3n + 1)^2 - 1^2\bigg] + 1$

. . $=\;(n^2 + 3n + 1)^2$