# proof about product of consecutive ints

• May 16th 2008, 07:26 PM
eniuqvw
proof about product of consecutive ints
how do you prove that the product of four consecutive integers is one less than a perfect square. I started by letting m be the lowest so m(m+1)(m+2)(m+3) + 1 should be a perfect square, i don't know how to show this algebraically :(
• May 16th 2008, 08:51 PM
Reckoner
Hi, eniuqvw!

Quote:

Originally Posted by eniuqvw
how do you prove that the product of four consecutive integers is one less than a perfect square. I started by letting m be the lowest so m(m+1)(m+2)(m+3) + 1 should be a perfect square, i don't know how to show this algebraically :(

Obviously, all that is needed is to show that $\displaystyle m\left(m+1\right)\left(m+2\right)\left(m+3\right)+ 1 = m^4+6m^3+11m^2+6m+1$ can be factored into the square of a quadratic in $\displaystyle m$. If you can factor this expression right away, then good for you! I, however, didn't have much luck (if anyone has a nice and simple way to factor this, let me know!), so I worked backwards:

The square of a quadratic will have the form

$\displaystyle \left(am^2 + bm + c\right)^2 = a^2m^4+2abm^3+\left(2ac+b^2\right)m^2+2bcm+c^2$

So, if $\displaystyle m^4+6m^3+11m^2+6m+1$ can indeed be factored as the square of a quadratic in $\displaystyle m$, then it should take that form. With this assumption, we have:

$\displaystyle \begin{array}{rcl} a^2 & = & 1\\ 2ab & = & 6\\ 2ac + b^2 & = & 11\\ 2bc & = & 6\\ c^2 & = & 1 \end{array}$

Solving this, we find that $\displaystyle \left(a,\,b,\,c\right) = \left(1,\,3,\,1\right)\text{ or }\left(-1,\,-3\,-1\right)$.

And, indeed, if you try expanding it yourself, you will see that

$\displaystyle \left(m^2 + 3m + 1\right)^2 = m^4+6m^3+11m^2+6m+1$

And this fact should give you what you need to complete your proof.
• May 16th 2008, 10:08 PM
Soroban
Hello, eniuqvw!

Quote:

Prove that the product of four consecutive integers is one less than a perfect square.
Let the four integers are: .$\displaystyle m,\:m+1,\:m+2,\:m+3$

We have: .$\displaystyle P \;=\;m(m+1)(m+2)(m+3)$

. . $\displaystyle P \;=\;[m(m+3)]\,[(m+1)(m+2)]$

. . . . $\displaystyle =\;[m^2 + 3m]\,[m^2 + 3m + 2]$

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

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