Help proving the product of any four consecutive integers is one less than a perfect square [duplicate]

Solution 1:

\begin{align} m(m+1)(m+2)(m+3)&=\left[(m)(m+3) \right] \left[ (m+1)(m+2)\right]\\ &=\left[m^2+3m \right] \left[ m^2+3m+2\right]\\ &=\left[(m^2+3m+1)-1 \right] \left[ (m^2+3m+1)+1\right]\\ &=(m^2+3m+1)^2-1 \end{align}

Solution 2:

If you have trouble approaching this, try some examples: Note that (for instance)

$$ 1 \times \color{red}{2 \times 3} \times 4 = 24 = 5^2-1 $$

$$ 2 \times \color{red}{3 \times 4} \times 5 = 120 = 11^2-1 $$

$$ 3 \times \color{red}{4 \times 5} \times 6 = 360 = 19^2-1 $$

and observe that

$$ 5 = \color{red}{2 \times 3} - 1 $$

$$ 11 = \color{red}{3 \times 4} - 1 $$

$$ 19 = \color{red}{4 \times 5} - 1 $$

So try seeing if

$$ m\color{red}{(m+1)(m+2)}(m+3) = [\color{red}{(m+1)(m+2)}-1]^2-1 $$

Solution 3:

You can take the product of the four consecutive integers as $$(m-1)m(m+1)(m+2)$$

Adding $1$ and simplifying, we get $$m^4+2m^3-m^2-2m+1$$ $$=m^4+m^2+1+2m^3-2m^2-2m$$ $$=(m^2+m-1)^2$$

Solution 4:

Intro: this is the business about taking the gcd of a polynomial and its derivative, in order to detect a repeated factor. In this case, the repeated factor just squares to give the original, so this is one of the simplest types. For this answer yesterday, How can one find the factorization $a^4 + 2a^3 + 3a^2 + 2a + 1 = (a^2 + a + 1)^2$ from scratch? I deliberately made up a more complicated situation. In that one, it was actually the cube of something.

$$ f = m^4 + 6 m^3 + 11 m^2 + 6 m + 1 $$ $$ f' = 4 m^3 + 18 m^2 + 22 m + 6 $$ $$ f'/2 = 2 m^3 + 9 m^2 + 11 m + 3 $$ $$ 2 f - m (2 m^3 + 9 m^2 + 11 m + 3) = 3 m^3 + 11 m^2 + 9 m + 2 $$ $$ 3(2 m^3 + 9 m^2 + 11 m + 3) - 2 (3 m^3 + 11 m^2 + 9 m + 2) = 5 m^2 + 15 m + 5. $$ With $$ 5 m^2 + 15 m + 5 = 5 (m^2 + 3m + 1) $$ we want $$ \gcd_{\mathbb Q}(2 m^3 + 9 m^2 + 11 m + 3,m^2 + 3m + 1 ). $$ However,, $$ 2 m^3 + 9 m^2 + 11 m + 3 = (2m+3)(m^2 + 3m + 1) $$ so the GCD is $m^2 + 3m + 1$ itself. This must be a repeat factor of the original $f,$ and we check that, in fact, $f = (m^2 + 3m+1)^2$