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Math Help - Question 2

  1. #1
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    Question 2

    A triangular number is the sequence:
    1,3,6,10,15,21,...

    A square number is the sequence:
    1,4,9,16,25,...

    Show that there are these two sequences contain infinitely many same numbers, i.e. 1 and 1.
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    Quote Originally Posted by ThePerfectHacker View Post
    A triangular number is the sequence:
    1,3,6,10,15,21,...

    A square number is the sequence:
    1,4,9,16,25,...

    Show that there are these two sequences contain infinitely many same numbers, i.e. 1 and 1.
    To clarify: You mean that the set of numbers that belong to both the triangular sequence and the square sequence has an infinite number of members?

    -Dan
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    Quote Originally Posted by topsquark View Post
    To clarify: You mean that the set of numbers that belong to both the triangular sequence and the square sequence has an infinite number of members?

    -Dan
    Yes
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  4. #4
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    I do not understand the last line of the solution?
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    Hello, TPHacker!

    Generally, with this type of problem we can generate solutions
    with a recurrence formula of the form:

    . . nth term .= .k(preceding term) (term before) ... for some constant k.


    In short, I "eyeballed" it . . .

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  6. #6
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    Beautiful job Soroban +rep+
    Never seen it done like that.
    Your own solution?

    I have modified (not really the right word) your recurrence relation into a different form:
    ---
    I always tried to show that the cubic sequence and triangular sequence only have a trivial solution i.e. 1. And it seems to be true for higher orders also but I was unable to prove that, you know how?
    Attached Thumbnails Attached Thumbnails Question 2-picture6.gif  
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  7. #7
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    Hello again, TPHacker!

    I got a different closed form for the recurrence.

    It's not my own solution; I learned the method from a few of my books.
    You may have seen something similar while exploring.

    Now that LaTeX is back, I'll revise this post.


    We have the recurrence: . f(n) \:=\:6\cdot f(n-1) - f(n-2)

    Assume that f(n) is exponential: . f(n) \:=\:X^n

    The equation becomes: . X^n \:=\:6X^{n-1} - X^{n-2}

    Divide by X^{n-2} and we have: . X^2 - 6X + 1 \:= \:0

    . . which has roots: . X \:=\:3 \pm 2\sqrt{2}


    We form a linear combination of the two roots:
    . . f(n) \:=\:A(3 + 2\sqrt{2})^n + B(3 - 2\sqrt{2})^n

    To solve for A and B, we use the first two values of the sequence:

    . . \begin{array}{cc}f(1) \:= \\ f(2) \:=\end{array} \begin{array}{cc} A(3 + 2\sqrt{2}) + (3 - 2\sqrt{2})\\ A(3 + 2\sqrt{2})^2 + (3 - 2\sqrt{2})^2\end{array}\begin{array}{cc}= \:1 \\ = \:6\end{array}

    Solve the system and get: . A \,= \,\frac{1}{4\sqrt{2}},\;\;B \,= \,-\frac{1}{4\sqrt{2}}

    Therefore: . f(n) \;= \;\frac{(3 + 2\sqrt{2})^n - (3 - 2\sqrt{2})^n}{4\sqrt{2}}


    By the way: . 3 \pm 2\sqrt{2} \:=\:(1 \pm \sqrt{2})^2

    . . so you can rewrite the formula if you like . . .

    Last edited by Soroban; October 23rd 2006 at 12:14 PM.
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    Besides for Soroban's solution.... I have 3 more (1 of which I am too lazy to post).

    Let t_n be the n-thtriangular number.

    Solution 1)
    If t_n is a square then t_{4n(n+1)} is a square. How you are suppopsed to see that, I have no idea.

    Solution 2)
    Following the start of Soroban;s solution we arrive at the square root of 8k +1, which is the "Pellian equation". Thus it has infinitely many solutions.
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