# Write as whole square

• Oct 6th 2010, 12:15 PM
razemsoft21
Write as whole square
\$\displaystyle n(n+1)(n+2)(n+3)+1\$
as whole square

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

My Try:
\$\displaystyle 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
• Oct 6th 2010, 12:17 PM
mr fantastic
Quote:

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

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

My Try:
\$\displaystyle 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
• Oct 6th 2010, 04:34 PM
razemsoft21
can you please show me the way you solve it ?
• Oct 6th 2010, 06:26 PM
mr fantastic
Quote:

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.
• Oct 6th 2010, 07:39 PM
Soroban
Hello, razemsoft21!

Quote:

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

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

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

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

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

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

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