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Math Help - proof about product of consecutive ints

  1. #1
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    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
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  2. #2
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    Hi, eniuqvw!

    Quote Originally Posted by eniuqvw View Post
    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 m\left(m+1\right)\left(m+2\right)\left(m+3\right)+  1 =<br />
m^4+6m^3+11m^2+6m+1 can be factored into the square of a quadratic in 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

    \left(am^2 + bm + c\right)^2 =<br />
a^2m^4+2abm^3+\left(2ac+b^2\right)m^2+2bcm+c^2

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

    \begin{array}{rcl}<br />
a^2 & = & 1\\<br />
2ab & = & 6\\<br />
2ac + b^2 & = & 11\\<br />
2bc & = & 6\\<br />
c^2 & = & 1<br />
\end{array}

    Solving this, we find that \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

    \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.
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  3. #3
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    Hello, eniuqvw!

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

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

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

    . . . . =\;[m^2 + 3m]\,[m^2 + 3m + 2]<br />

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

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

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