1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor

    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

    Attached Files:

  8. Jan 5, 2010 #7


    User Avatar
    Science Advisor
    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


    User Avatar
    Homework Helper

    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 .
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook