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Thread: Four consecutive integers

  1. #1
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    Four consecutive integers

    Prove that the product of four natural consecutive numbers is equal to the difference of the squares of two natural numbers.
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    Re: Four consecutive integers

    Quote Originally Posted by louis33 View Post
    Prove that the product of four natural consecutive numbers is equal to the difference of the squares of two natural numbers.
    If $P(n)=(n)(n+1)(n+2)(n+3)$ can you show that $P(1)=5^2-1^2~?$
    Can you finish?
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    Re: Four consecutive integers

    Yeah, but I'm supposed not to give any values to that n but to rather solve in a generalizing way.
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    Re: Four consecutive integers

    $n(n+1)(n+2)(n+3)=\left(n^2+3n+1\right)^2-1$
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    Re: Four consecutive integers

    Quote Originally Posted by louis33 View Post
    Yeah, but I'm supposed not to give any values to that n but to rather solve in a generalizing way.
    WHAT are you saying? Do you know about induction?
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    Re: Four consecutive integers

    Quote Originally Posted by Idea View Post
    $n(n+1)(n+2)(n+3)=\left(n^2+3n+1\right)^2-1$
    How did you cook that up?
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    Re: Four consecutive integers

    Quote Originally Posted by rodders View Post
    How did you cook that up?
    HERE is one way.
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    Re: Four consecutive integers

    Quote Originally Posted by Plato View Post
    HERE is one way.
    Ha!
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    Re: Four consecutive integers

    $(a^2-d^2)n^4 + (2ab-2de)n^3 + (2ac+b^2-2df-e^2)n^2+(2bc-2ef)n+c^2-f^2 = \underbrace{(an^2+bn+c)^2-(dn^2+en+f)^2}_{\text{A difference of squares}} = n(n+1)(n+2)(n+3) = n^4+6n^3+11n^2+6n$

    Equate coefficients on the far left with the far right:
    Eq 1: $a^2-d^2 = 1$
    Eq 2: $2ab-2de = 6$
    Eq 3: $2ac+b^2-2df-e^2=11$
    Eq 4: $2bc-2ef = 6$
    Eq 5: $c^2-f^2 = 0$

    Factoring Eq 1 gives:
    $(a+d)(a-d) = 1$
    The only factors of 1 are 1 and 1 or -1 and -1, so we have $a+d = a-d$. This gives us $d=0, a=\pm 1$. With no loss of generality, we can assume $a=1$. This process follows similarly for $a=-1$, but you will not wind up with a natural number for the first number. You will get negative the results you find for a,b,c. You can try it out yourself if you want. For now, assume $a=1$. Plug that in to Eq 2:
    $2(1)b-2(0)e = 6$ implies $b=3$ and we are not sure about $e$. Let's plug in what we know to the last three equations:

    Eq 3.1: $2(1)c+3^2-2(0)f-e^2=11$ implies $2c-e^2=2$. Since $-e^2 \le 0$, it must be that $2c \ge 2$ implies $c\ge 1$.
    Eq 4.1: $2(3)c-2ef=6$ implies $3c-ef=3$. Again, we have $c\ge 1$.
    Eq 5.1: $c^2-f^2=0$ implies $c=\pm f$.

    Case 1: $c=f$
    Plug into Eq 4.1:
    $3c-ec=3$ implies $c(3-e)=3$. Therefore, $c=1$ or $c=3$ (note: we already figured out that $c$ must be positive from Eq 3.1). Then $3-e = 1$ or $3-e= 3$ are the only possibilities. Here are the possibilities:
    $c=f=1; e=0$. Plugging into Eq 3.1 gives: $2(1)-0^2 = 2$, which is true. This gives: $(n^2+3n+1)^2-(0n^2+0n+1)^2 = (n^2+3n+1)^2-1$
    $c=f=3; e=2$. Plugging into Eq 3.1 gives: $2(3)-2^2=2$, which is true. This gives $(n^2+3n+3)^2-(0n^2+2n+3)^2 = (n^2+3n+3)^2-(2n+3)^2$
    Case 2: $c=-f$
    Plug into Eq 4.1:
    $3c+ec=3$ implies $c(3+e)=3$. Therefore, $c=1$ or $c=3$ and $3+e=3$ or $3+e=1$. This gives:
    $c=1;f=-1;e=0$ which gives $(n^2+3n+1)^2-(0n^2+0n-1)^2$. But, the latter one is not a natural number, so we can throw this possibility out.
    $c=3;f=-3;e=-2$ which gives $(n^2+3n+3)^2-(0n^2-2n-3)^2$. Again, this is never a natural number, so we can throw this possibility out, as well.

    That gives us two possible differences of two squares of natural numbers:
    $(n^2+3n+1)^2-1$
    $(n^2+3n+3)^2-(2n+3)^2$

    If we wanted, we could have used higher exponents than just $n^2$. But, that would make this far too complicated.
    Last edited by SlipEternal; Jan 26th 2018 at 12:39 PM.
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    Re: Four consecutive integers

    Note: . . . The content of my post was beaten in time by SlipEternal. I looked on my
    page for somewhere to delete it so SlipEternal's post would be the last to show, but
    I could not find any delete option.


    Quote Originally Posted by Idea View Post
    $n(n+1)(n+2)(n+3)=\left(n^2+3n+1\right)^2-1$
    Aside from changing all of the signs of the trinomial inside the parentheses, that is not the
    only way. Here is another way:

    $n(n + 1)(n + 2)(n + 3) = (n^2 + 3n + 3)^2 - (2n + 3)^2$

    For instance, when n = 1, $ \ P(1) = 24 = 7^2 - 5^2.$

    I looked at $ \ (an^2 + bn + c)^2 - (dn + f)^2 = n^4 + 6n^3 + 11n^2 + 6n = n(n + 1)(n + 2)(n + 3) \ \ $ **

    I equated coefficients on the different degreed n variables and solved a system of equations.

    There are many solutions.

    Taking the non-negative integer solutions gives:

    $(n^2 + 3n + 1)^2 - (1)^2$

    $(n^2 + 3n + 3)^2 - (2n + 3)^2$


    ** I used "f" instead of "e" because of the other common use of e.
    Last edited by greg1313; Jan 26th 2018 at 12:43 PM.
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  11. #11
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    Re: Four consecutive integers

    So there are only two ways of expressing the difference of squares?
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    Re: Four consecutive integers

    For a given $n$ let $p=n(n+1)(n+2)(n+3)$

    For any divisor $r$ of $p$ , $r^2<p$, and both $r$ and $p/r$ even, let

    $a=\frac{r}{2}+\frac{p}{2r}$ and

    $b=\frac{p}{2r}-\frac{r}{2}$

    We then have
    $p=a^2-b^2$

    so in general more than two solutions
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    Re: Four consecutive integers

    Ok. so its possible to express p in more than two ways ( as a difference of two squares)?
    I can't seem to find a numerical example of this? (yet!)
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    Re: Four consecutive integers

    $n=2$
    $p=120$

    $120=11^2-1^2=13^2-7^2=17^2-13^2=31^2-29^2$
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    Re: Four consecutive integers

    Quote Originally Posted by Idea View Post
    For a given $n$ let $p=n(n+1)(n+2)(n+3)$

    For any divisor $r$ of $p$ , $r^2<p$, and both $r$ and $p/r$ even, let

    $a=\frac{r}{2}+\frac{p}{2r}$ and

    $b=\frac{p}{2r}-\frac{r}{2}$

    We then have
    $p=a^2-b^2$

    so in general more than two solutions
    How have you decided on 'let a and b take the form above?'
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