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Tricky question, need help

  1. Jun 15, 2004 #1
    What number when squared and either increased or decreased by 5 gives a number that is a perfect square?
     
  2. jcsd
  3. Jun 15, 2004 #2

    AKG

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    The number we're looking for, let's call it [itex]a[/itex]. Let's call the perfect square [itex]b^2[/itex]. So:

    [tex]a^2 \pm 5 = b^2[/tex]
    [tex]a^2 - b^2 = \mp 5[/tex]
    [tex](a - b)(a + b) = \mp 5[/tex]

    Now, if a > b, then a - b > 0, and a + b > 0, therefore (a - b)(a + b) > 0, and thus is not equal to -5, so it's +5. If a < b, then it's -5. We can basically see that whether we choose a > b or b > a, it's not really going to matter. In other words, if we chose a > b, we would find a certain value for a and a certain value for b. If we chose b > a, we would find the same values, but just swapped around, but it really doesn't matter which one we call a and which one we call b. So, let's just go with a > b:

    [tex](a - b)(a + b) = 5[/tex]

    Now, 5 is a prime number, and the only two numbers that divide 5 are 1 and 5, so we can conclude a - b = 1, and a + b = 5. Solve, and we get a = 3, b = 2. As you can see, [itex]3^2 - 5 = 2^2[/tex]. If we had gone the other way, and gotten a = 2, b = 3, we would simply have [itex]2^2 + 5 = 3^2[/itex] and since the question said that we can either add or subtract 5, either answer should be acceptable.
     
  4. Jun 15, 2004 #3
    well, if you list the squares of the positive interger set you might be able to see things easier.

    1, 2, 3, 4, 5, 6, 7
    1, 4, 9, 16, 25, 36, 49

    that doesnt help much except to see that there isnt a square that is +/- 10 of another square. i dont see how any number can into this definition, (x*x)+-5=c*c where c is an interger. (x*x)+5 and (x*x)-5 would mean the two perfect squares would lie 10 intergers apart, and there arent any 2 consecutive perfect squares that lie an even number apart. a perfect square n^2 predicts the next perfect square more simply as n^2 + n + n + 1. you can visualize a square with n units per side. if you increase the size of the square by one, you can see how you are adding a strip of n+1 and n to the original n*n square size. based on this you would have to find non consecutive perfect squares. 1,4,9,16 shows us that 1 and 9 are 8 apart, and 4 and 16 are 12 apart. there is no pair inbetween that satisfies the original requirement.

    if i am misunderstanding your question...

    one possible interpretation of your wording is that a number squared and either increased or decreased means the number need not be both increased and decreased, but only one or the other. that would mean x^2+5=c^2 with c being the interger.
    lets plug in anything for c equals 3. you get x^2+5=9 x^2=4, x=2. this satisfies the main equation, but i dont see how -1 could be a perfect square. you can choose these because you know that 4 - 9 = 5. 2^2 - 3^2 = 5.


    if you interpret "when squared, and either increased or decreased by 5" to mean that the number is increased or decreased by 5 before the squaring... then you have (x+5) or (x-5) when either are taken to the 1/2 power (that means square root) equals a perfect square.

    am i misunderstanding something? is the problem solvable?
     
  5. Jun 16, 2004 #4
    One possible solution is: X = sqrt(29)/2

    Check: X2 = 29/4

    And X2 - 5 = 29/4 - 5 = 9/4 = (3/2)2 ... a square.

    And X2 + 5 = 29/4 + 5 = 49/4 = (7/2)2 ...

    I checked today and this is the right solution now can anyone find the equation so that any number will work in place of 5,-5 like 13,-13. It is a one varable equation.
     
  6. Jun 16, 2004 #5
    depends on what you meant by perfect square( i thought it was an integer squared)
     
  7. Jun 16, 2004 #6
    The answer I gave is one correct answer there are many to this question from what my teacher says. I still need to find the formula.
     
  8. Jun 16, 2004 #7

    AKG

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    That's very strange. [itex](3/2)^2[/itex] is not a perfect square. I wonder now what you mean by perfect square. A perfect square is in this context should be the square of an integer, but clearly the "perfect square" you've given is the square of a rational. Are squares of irrationals also acceptable, because if so almost every number would be acceptable. You really need to clarify some things in light of your new example

    "What number when squared and either increased or decreased by 5 gives a number that is a perfect square?"

    What kind of limitations do we have on this number. Can it be a complex number? Must it be a real number, must it be positive? The answer you've provided in your example is irrational; is it safe to assume this is acceptable? And please define what you / your teacher mean by perfect square.
     
  9. Jun 16, 2004 #8
    When refering to numbers, the only accepted meaning of "perfect square" is a number where the square root is a positive integer. It is very poor practice to teach people a non-standard meaning of such a term.
     
  10. Jun 17, 2004 #9

    Gokul43201

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    If your difference between squares is some N, solutions are

    [tex] a = \frac {x + N/x} {2} [/tex]
    [tex] b = \frac {x - N/x} {2} [/tex]

    for all x (in the domain of your choice, since this is not clear yet)

    Integral pair solutions exist for all differences that are odd or multiples of 4 (and for no others).

    Edit : Misunderstood question. Actually, need solutions for the requirement that adding "as well as" subtracting N give rational squares.

    Will rethink.
     
    Last edited: Jun 17, 2004
  11. Jun 17, 2004 #10

    Gokul43201

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    Since integral squares are separated by odd numbers or multiples of 4(add multiples of 8 to convert from necessary to sufficient), the only way to get a difference of say, 10 (see abertram's frustration) is to find a functioning difference that is a multiple of 10 (such as 40, between 9 and 49) - we know that this exists, since 40 is a multiple of 8 - and divide by 4.

    Hence, 49/4 and 9/4 are the outer squares and the solution is the square root of the mean, ie : sqrt(29)/2
     
  12. Jun 17, 2004 #11
    i always thought perfect squares had to be squares of integers? :confused:
     
  13. Jun 17, 2004 #12
    They are, seems there is a misunderstanding here.
     
  14. Jun 17, 2004 #13

    Gokul43201

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    Another solution is found by looking for a difference of 160 between integer squares. This is found for 9 and 169. Hence 9/16 and 169/16 are the outer squares. So sqrt(89)/4 is another solution.

    An infinite number of such solutions can be found, because there are an infinite number of multiples of GCD(10,8), such that those multiples are 10 times a perfect square.

    eg : you can always find squares that have differences of 40, 160, 360, 640, ...10*(2n)^2 and they are ({9,49},{36,196},{81,441},{144,784}...{9n^2,49n^2})

    So that's your general solution...at least for 10. (maybe)You can generalize it further for any N.
     
    Last edited: Jun 17, 2004
  15. Jun 17, 2004 #14

    Gokul43201

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    These are not the only solutions for D = 10. Other pairs like {9,169}, {36,676}, etc. can be found of the form {9n^2, 169n^2} and so on...
     
  16. Jun 17, 2004 #15
    Well thanks for all the help everyone. When I get the answer at the end of the semester I will share it with all of you, it has stumped me and at the same time made me a little mad....lol. According to the teacher my answer is correct, but I dont know how concidering (29)/2 is not a perfect square. Again thanks alot everyone!!
    I am anxious to figgure this out.
     
  17. Jun 17, 2004 #16

    Gokul43201

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    I gave you the answer(s) in my previous posts.
     
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