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Prove that the square of an odd integer

  1. Sep 9, 2004 #1
    I'm not a very logical person, and I would hardly consider math a strength so, I'm stuck with these proofs:

    1. Prove that every positive integer, ending in 5 creates a number that when squared, ends in 25

    2. Prove that if n is an even positive integer, then n^3 - 4n is always divisible by 48.

    3. Prove that the square of an odd integer is always of the form 8k + 1 , where k is an integer.

    Thank you. :wink:
  2. jcsd
  3. Sep 9, 2004 #2


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    In all these problems you know something about the form of the integers being considered, use this information.

    For example in Q2, n is even so you can write it as n=2m for some positive integer m. Substitute this into n^3-4n and try to show it's divisible by 48, irregardless of the value of m (do you know modular arithmetic yet?).

    Same deal will work for Q3, except your number is odd here.

    For Q1, can you think of a simple general form for numbers ending in 5? I can't think of a good hint off hand that doesn't give too much away...
  4. Sep 9, 2004 #3
    I've never studied the modular arithmetic yet. Unless you are referring to a arithmetic sequence?

    Anyways, working ahead on Q2, I've gotten from:

    n^3 - 4n



    2m^3 - 4 (2m)

    2m^3 - 8m

    2m (m^2 - 4)

    2m (m + 2) (m - 2)

    Not sure where to go from there?
  5. Sep 9, 2004 #4


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    That should be [itex]8m^3-8m[/itex] which, by the way can be expressed as [itex]8m(m-1)(m+1)[/itex]. :-)
  6. Sep 9, 2004 #5
    I really have no clue what your doing but doesnt that factor to 8m(m-1)(m+1)?

    I dont understand what we are trying to do here...

    (2)^3-4(2) = 0
    is 0 divisible by 48??? I guess so, but then again.. everything is divisible by everything, its just not guaranteed to give a whole number as an answer.
  7. Sep 9, 2004 #6


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    Careful with those substitutions, brackets are easy to miss:)

    Now, you can clearly see that 8 divides [itex]8m(m-1)(m+1)[/itex], because of the 8 in front. What can you say about the [itex]m(m-1)(m+1)[/itex] part? Is it even? odd? divisibly by what? Try subbing in a few values of m (say 1 to 5) and see if anything jumps at you.

    ps. Modular arithmetic is different from arithmetic sequences, and can give a handy way of dealing with divisibility kind of problems. It's not necessary here though, so don't worry about it.

    pps. 0 is divisible by everything, in the integer sense of "divisible". we say a is divisible by b if there exist c where a=bc (a,b, c all integers). In the case of a=0 and b=anything, take c=0.
  8. Sep 9, 2004 #7


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    schmoe: BTW - you don't need modulo arithmetic.

    musky ox: Divisibility in math has a specific meaning - namely an integer is divisible by another integer if there is no remainder.
  9. Sep 9, 2004 #8


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    For the 5 proof, this is how I would go about it (I do this for ending in 5, and NOT ending in 25, to give you something to expand on):

    We have a number ending with 5 which has another digit in it, x is the other digit, and y it's position (with 0 being the position of the 5, so y > 0, and x and y are contained within Z). This way we can show if a digit is in any position to the left of the 5, it won't affect the final result.

    (x*10^y+5)^2 = (x*10y)^2 + 2*x*10^y*5 + 5^2 = 100*(xy)^2 + 10*x*10^y + 25

    Now, we only want the last digit so we mod by 25, now:
    (x + y) mod z = ((x mod z) + (y mod z)) mod z
    (x*y) mod z = ((x mod z)(y mod z)) mod z

    *by the way, this equation really spreads out, and I don't know latex. The underlined abd bolded portions signify work being done.

    (100*(xy)^2 + 2x*10^2 + 25) mod 10
    = [([100*(xy)^2] mod 10) + ([10*x*10^y] mod 10) + (25 mod 10)] mod 10
    = ([([100 mod 10]*[(xy)^2 mod 10]) mod 10] + ([10 mod 10] * [(x*10^y) mod 10] mod 10) + 5) mod 10
    = ([(0 * [(xy)^2 mod 10]) mod 10] + (0 * [(x*10^y) mod 10] mod 10) + 5) mod 10
    = (0 + 0 + 5) = 5

    By the way.. I just guessed at the mod rules, and if they don't stand the entire proof is false.
    Last edited: Sep 9, 2004
  10. Sep 9, 2004 #9


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    (10k + 5)^2 = 100k^2 + 100k + 25 = 100(k^2 + k) + 25

    (2n)^3 -4*(2n) = 8n^3 - 8n = 8n(n-1)(n+1) = 8(n-1)n(n+1)
    One of (n-1), n, (n+1) must be divisible by three, and at least one of them must be even, therefore 48|(2n)^3 -4*(2n)

    (2n + 1)^2 = 4n^2 + 4n + 1 = 4n(n + 1) + 1
    Either n or (n+1) is even, therefore (2n + 1)^2 = 8k+1
  11. Sep 10, 2004 #10


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    A lot of really complicated responses and then CTS gives the obvious way to do them! (Just before I could!)
  12. Sep 10, 2004 #11


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    I considered going down that route, but then I realised that someone could argue that the digit wasn't in the "ten" position, and thus the proof didn't cover it. Much shorter than my work though! :blushing:
  13. Sep 10, 2004 #12
    3: (2m+1)^2 = 4m^2+4m+1
    4m^2+4m+1 = 4m(m+1)+1
    Now we have to cases, if m is even let m=2n then we get
    8n(2n+1)+1 which is of the form 8n*(2n+1)+1. If m is odd we have m=2u+1 which yields
    4*(2u+1)*(2u+2) + 1 = 8*(2u+1)*(u+1)+1 which is of the form 8k+1.
  14. Sep 10, 2004 #13


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    I may have been more complicated, but my hope was to kick Yodlums down the path to understand how to actually think about solving these sort of problems. I figured that would be much more instructive than to just hand out the solutions (teach a man to fish and all that jazz). Oh well.

    Tide-I know modular arithmetic isn't necessary (see my post above yours) but it's often taught close by divisibility. If that was the case here, it would have made sense to use that language where appropriate.
  15. Sep 10, 2004 #14


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    Schmoe - yes I reread your post and realize I had read it too fast the first time! Thanks for the clarification.
  16. Sep 12, 2004 #15
    The first one i thought was pretty simple, i'm taking the same course as you right now (i'm guessing), and probably have the same textbook too (Goemetry and Discrete Mathematics)...this is how i did #1 for my homework :zzz:

    1. Prove that every positive integer, ending in 5 creates a number that when squared, ends in 25

    Okay, listen to what the question is asking you to prove, it says prove that for every POSITIVE INTEGER ending in 5 and then SQUARED, generates a number that ends in 25. Think about a formula that would generate a number ending in 5 for whatever positive integer you input, i used this..

    Let n = 5(2x+1) (every odd number multiplied by 5 will end in 5 :smile:)
    = 10x+5 (expand)

    We now have a number that ends in 5 no matter what we plug in (positive integers)...now it has to be squared doesent it?...cause remember what the original question said...so we simply square our "formula".

    Let P(n) = n^2
    = 100x^2 + 100x +25

    Now this is where you have to think, the 100's in the equation are gonna make numbers ending in 00 no matter what so you can conclude that...

    = 00 + 00 + 25
    = 25

    Theres number one, i realize that other people already posted their answer to this, but i decided to show it in a more simple manner.
  17. Sep 12, 2004 #16


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    Me too, shmoe, I was really referring to the posts that answered in terms of "mod" which was clearly too technical for the original question.
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