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Homework Help: Product of two consecutive integers

  1. Jan 4, 2010 #1
    1. The problem statement, all variables and given/known data

    Prove that [tex]n^2+n[/tex] is even. Where n is a positive integer.

    2. Relevant equations


    3. The attempt at a solution

    [tex]n^2+n = n(n+1)[/tex]

    One of which must be even, and therefore the product of 2 and an integer k.

    [tex]n = 2k, \left \left 2*(k(n+1))[/tex]


    [tex]n+1 = 2k, \left \left 2*(n*k)[/tex]

    Is there a better way of doing this? I read this is not an inductive proof; what would this entail?
    Many thanks in advance.
  2. jcsd
  3. Jan 4, 2010 #2
    This looks absolutely fine. No need for induction at all.
  4. Jan 4, 2010 #3


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    As rochfor1 said, that is a perfectly good proof and simpler than "proof by induction".

    But since you ask, here goes:

    If n= 1, then [itex]n^2+ n= 1^2+ 1= 2[/itex] which is even.

    Now, suppose that [itex]k^2+ k[/itex] is even and look at [itex](k+1)^2+ (k+1)[/itex]
    [itex](k+1)^2+ (k+1)= k^2+ 2k+ 1+ k+ 1[/itex][itex]= (k^2+ 2k)+ 2k+ 2[/itex]. By the induction hypothesis, [itex]k^2+ k[/itex] is even and so [itex]k^2+ k= 2m[/itex] for some integer m (that is the definition of "even") so [itex](k+1)^2+ (k+1)= (k^2+ 2k)+ 2k+ 2[/itex][itex]= 2m+ 2k+ 2= 2(m+k+1)[/itex]. Since that is "2 times an integer", it is even.

    Having proved that the statement is true for 1 and that "if it is true for k, it is true for k+1", by induction, it is true for all positive integers.

    Yet a third way: If n is a positive integer, it is either even or odd.

    case 1: n is even. Then n= 2m for some integer m. [itex]n^2= 4m^2[/itex] so [itex]n^2+ n= 4m^2+ 2m= 2(m^2+ m)[/itex]. Since that is 2 times an integer, it is even.

    case 2: n is odd. Then n= 2m+ 1 for some integer m. [itex]n^2= (2m+1)^2= 4m^2+ 4m+ 1[/itex] so [itex]n^2+ n= 4m^2+ 4m+ 1+ 2m+ 1[/itex][itex]= 4m^2+ 6m+ 2= 2(2m^2+ 3m+ 1)[/itex]. Again that is 2 times an integer and so is even.
  5. Jan 5, 2010 #4
    Many thanks rochfor1 and HallsofIvy!
  6. Jan 5, 2010 #5
    Simple reasoning is needed.
    For two consecutive integer , one will be even and another will be odd.
    [tex]n\equiv0(mod 2)[/tex]
    [tex]n+1\equiv1(mod 2)[/tex]

    We may now conclude that [tex]n(n+1)\equiv0(mod 2)[/tex] and means that n²+n is even.
  7. Jan 5, 2010 #6

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  8. Jan 5, 2010 #7


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    Homework Helper

    If n is odd, n^2 is odd. If n is even, n^2 is even. odd+odd=even and even+even=even.
  9. Jan 6, 2010 #8


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    OP: Why do you want a proof by induction? The proof you gave is simple and beautiful.
  10. Jan 6, 2010 #9


    Staff: Mentor

    What you say is true, but you can't ascertain which of them will be even and which will be odd. In what follows, you are assuming that n is even and n+1 is odd. That is one of two possible cases, so you work is not complete.
  11. Jan 7, 2010 #10
    yes that is true. for the other case , it can be proved analogously .
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