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Difference of two squares

  1. Sep 12, 2009 #1
    1. The problem statement, all variables and given/known data

    1) Find all pairs of natural numbers whose squares differ by 75.

    2) Find all pairs of natural numbers whose squares differ by 79.

    3) Prove that there can only be 1 pair of numbers with a prime number difference


    2. Relevant equations

    none

    3. The attempt at a solution

    from common sense i can conclude that the answer to question 1 is: 5 and 10. But i need an equation of some sort to get question 2.

    and im not exactly sure what question 3 is asking.
     
  2. jcsd
  3. Sep 13, 2009 #2
    hint for part 3:
    [tex]n=x^{2}-y^{2}=(x+y)(x-y)[/tex]
    where n is some prime number.

    Does that make sense?

    Then try part 2. Isn't 79 a prime number?
     
    Last edited: Sep 13, 2009
  4. Sep 13, 2009 #3
    mhm, i had already figured that out. but i solved for x and then substituted it into the equation, to get the number and i got 79 = 79, or 75 = 75, depending on the problem i was working out.
     
  5. Sep 13, 2009 #4
    What is the definition of a prime number? Look at the part on the right.
     
  6. Sep 13, 2009 #5
    [tex] 79 = (x+y)(x-y)[/tex]

    but how do i solve for one variable so i can substitute. when i do, it cancels each other out.

    EDIT: would it be a trick question? since primes can only be multiplied by 1 and their self. whereas 75 is not prime (3*25).
     
  7. Sep 13, 2009 #6

    HallsofIvy

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    You know that [itex]x^2- y^2= (x- y)(x+ y)= 79[/itex]. Further you know that x and y are positive integers so x- y and x+ y are also whole numbers. How many pairs of positive integers are there that multiply to give 79? Set one of such a pair equal to x- y, the other equal to x+ y and solve the two equations.

    How many pairs of positive integers are there that multiply to give 75? No, it's not a "trick" question.
     
  8. Sep 13, 2009 #7
    theres only 1 pair that can make 79 since its prime. 1 and 79.

    if you set 1 and 79 as y and x you get

    [tex]79 = (79+1)(79-1)[/tex]

    [tex]79 = (80)(78)[/tex]

    which isnt correct.

    I know there is something that you are telling me that i just dont understand.
     
  9. Sep 13, 2009 #8
    [tex]x^2-y^2=79[/tex]

    [tex]y=\sqrt{x^2-79}[/tex]

    What is the range of the function y?
     
  10. Sep 13, 2009 #9
    i tried that, when i substituted it for y i got:

    [tex] 79 = x^2 -(\sqrt{x^2 - 79})^2 [/tex]

    [tex]79 = x^2 - x^2 -79[/tex]

    [tex]79=-79[/tex]
     
  11. Sep 13, 2009 #10
    You're not quite understand what was said. Don't set 79=x and y=1. Set (x-y)=79 and (x+y)=1, then the reverse, (ie. (x-y)=1 and (x+y)=79).
     
  12. Sep 13, 2009 #11
    OH, because we are finding the squares we need to find the what numbers equal the factors, i get it.

    [tex]x-y=1[/tex]
    [tex]y = -x - 1[/tex]

    [tex]x+y = 79[/tex]
    [tex]y = -x + 79[/tex]

    [tex]x+y = 1[/tex]
    [tex]y=-x + 1[/tex]

    [tex]x-y=79[/tex]
    [tex]y=x-79[/tex]

    but what do i do with these values?
     
  13. Sep 13, 2009 #12
    You know have 2 sets of systems of equations each solvable.
     
  14. Sep 13, 2009 #13
    and when you solve it you get 40 and 39.

    and [tex] 40^2 - 39^2 = 79 [/tex]

    AWESOME!

    and also

    [tex] 79 = (40-39)(40+39)[/tex]

    PERFECT. Thanks so much!
     
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