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Math Help - Perfect squares of n and n+99

  1. #1
    Senior Member Mukilab's Avatar
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    Perfect squares of n and n+99

    How many positive integer solutions are there n such that both n and n+99 are perfect squares?

    I believe there is only 1 solution.

    Any tips on how to tackle this? I haven't really come up with a method or a formula to calculate it, simply gone through the first 20 squares, and I doubt there will be anything after that

    Please help out, thanks
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  2. #2
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    i tried the first couple on paper only the number 1 is working. so i guess that it
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  3. #3
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    Quote Originally Posted by Mukilab View Post
    How many positive integer solutions are there n such that both n and n+99 are perfect squares?

    I believe there is only 1 solution.

    Any tips on how to tackle this? I haven't really come up with a method or a formula to calculate it, simply gone through the first 20 squares, and I doubt there will be anything after that

    Please help out, thanks
    I can't provide you with an equation or algorithm but I used an Excel table and have found: <br />
\begin{array}{cccc}nr&sqr&NR&SQR\\1& 1& 10& 100\\<br />
15& 225& 18& 324\\<br />
49& 2401& 50& 2500\end{array}

    Now the difference between the squared numbers is 1. So there can't be any other pair of numbers which satisfy the given conditions.
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  4. #4
    Senior Member Mukilab's Avatar
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    How does 49 satisfy the conditions?

    You must have misread my question

    take n as 1
    1^2 is a perfect square

    (1+99)^2 = 100 = 10^2


    49 does not suit this

    49+99=148

    15 does not suit this
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  5. #5
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by earboth View Post
    I can't provide you with an equation or algorithm but I used an Excel table and have found: <br />
\begin{array}{cccc}nr&sqr&NR&SQR\\1& 1& 10& 100\\<br />
15& 225& 18& 324\\<br />
49& 2401& 50& 2500\end{array}

    Now the difference between the squared numbers is 1. So there can't be any other pair of numbers which satisfy the given conditions.
    n \in \{1, 225, 2401\} is correct. The answer can be obtained with a Diophantine equation.

    n = k^2

    n+99 = m^2

    k^2-m^2+99 = 0

    See Dario Alpern's quadratic Diophantine solver and plug in a=1, c=-1, f=99 and select "step-by-step".

    Edit: Actually, the steps shown for this particular problem aren't as illuminating as they often are. Not sure the best way to work it out on paper.

    Edit 2: See Soroban's post.. seems so obvious now.
    Last edited by undefined; July 7th 2010 at 11:04 AM.
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  6. #6
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    Quote Originally Posted by Mukilab View Post
    How does 49 satisfy the conditions?

    You must have misread my question

    take n as 1
    1^2 is a perfect square

    (1+99)^2 = 100 = 10^2


    49 does not suit this

    49+99=148

    15 does not suit this
    Well, as usual, I've made your question a bit to difficult.

    So take

    225 +99 = 324 = 18^2

    2401+99 = 2500 = 50^2
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  7. #7
    Senior Member Mukilab's Avatar
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    that's taking it as n^2+99 but the actual question is n+99
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  8. #8
    MHF Contributor undefined's Avatar
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    By the way, I worked it out by brute force initially as well using PARI/GP which is a great tool (and hopefully you won't get spoiled by it). I wrote

    Code:
    f(n)=if(sqrt(n)==floor(sqrt(n)),1,0)
    for(i=0,10000,if(f(i)&&f(i+99),print(i)))
    Could also have used

    Code:
    f(n)=if(floor(sqrt(n))^2==n,1,0)
    which ought to be a bit more efficient.
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  9. #9
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by Mukilab View Post
    that's taking it as n^2+99 but the actual question is n+99
    There are three solutions.

    n=1=1^2 where n+99=100=10^2

    n=225=15^2 where n+99=324=18^2

    n=2401=49^2 where n+99=2500=50^2
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  10. #10
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    Hello, Mukilab!

    How many positive integers n are there
    such that both n \text{ and }n+99 are perfect squares?

    The answer is 3.


    We have two squares that differ by 99: . a^2 - b^2 \:=\:99

    So we have: . (a+b)(a-b) \:=\:\begin{Bmatrix}99\cdot1 \\ 33\cdot 3 \\ 11\cdot 9 \end{Bmatrix}


    Solve the three systems of equations:


    . . \begin{array}{ccc}a+b &=& 99 \\ a-b &=& 1 \end{array} \quad\Rightarrow\quad (a,b) \:=\:(50,49) \quad\Rightarrow\quad n = 2401


    . . \begin{array}{ccc}a+b &=& 33 \\ a-b &=& 3\end{array} \quad\Rightarrow\quad (a,b) = (18,15) \quad\Rightarrow\quad n = 225


    . . \begin{array}{ccc}a+b &=& 11 \\ a-b &=& 9 \end{array} \quad\Rightarrow\quad (a,b) = (10,1) \quad\Rightarrow\quad n = 1
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