JB
Lv 7
JB asked in Science & MathematicsMathematics · 1 decade ago

Can you find all positive integers such that this sum is a square?

Let S(n) = 1^5 + 2^5 + 3^5 + ... + n^5. Find all positive integers, n, such that S(n) is the square of an integer. Prove your answer.

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  • Duke
    Lv 7
    1 decade ago
    Best answer

    Excellent work by Steiner1745, Falzoon and RH above! They have shown the solutions, RH has suggested the best approach.

    Reading the question too late, I can only submit a proof using the theory of Pell's Equation (link in Sources below). As other answerers note, we should take care the expression

    (2n² + 2n - 1)/3 to be a perfect square, let

    2n² + 2n - 1 = 3y², or 4n² + 4n - 2 = 6y², or (2n + 1)² - 3 = 6y²,

    or, denoting x = 2n + 1 we reach the Diophantine equation:

    (1) x² - 6y² = 3 /generalized Pell's equation/

    Let's consider along with (1) also the standard Pell's equation:

    (2) ξ² - 6η² = 1

    According the theory if we have a fundamental solution (ξ_{1}, η_{1}), then all solutions of (2) are given by formulas /k=0, 1, 2, . ./:

    ξ_{k} = ((ξ_{1} + η_{1}*√6)^k + (ξ_{1} - η_{1}*√6)^k) / 2;

    η_{k} = ((ξ_{1} + η_{1}*√6)^k - (ξ_{1} - η_{1}*√6)^k) / (2√6);

    /for k=0 we get the trivial solution ξ_{0} = 1, η_{0} = 0/

    In our case the fundamental solution is ξ_{1} = 5, η_{1} = 2, so

    ξ_{k} = ((5 + 2√6)^k + (5 - 2√6)^k) / 2;

    η_{k} = ((5 + 2√6)^k - (5 - 2√6)^k) / (2√6) for k = 0, 1, 2, 3, . .

    Further, fundamental solution of (1) is x' = 3, y' = 1 and the solutions of (1) are given by:

    x_{k} = x' * ξ_{k} + 6y' * η_{k} = 3ξ_{k} + 6η_{k};

    y_{k} = x' * η_{k} + y' * ξ_{k} = 3η_{k} + ξ_{k}

    Easily can be seen that both ξ_{k} and x_{k} are odd, finally

    (3) n = (x_{k} - 1)/2 yields natural number for any k = 0, 1, 2, 3, . . - the required result.

    So the explicit non-recursive expression (much more inconvenient) is:

    n = 3((5 + 2√6)^k + (5 - 2√6)^k) / 4 + √6((5 + 2√6)^k - (5 - 2√6)^k) / 4 - 1/2

    Here are some values:

    k | ξ_{k} η_{k} | x_{k} y_{k} | n

    - - - - - - - - - - - - - - - - - - - - - - -

    0 | . . 1 . . 0 . | . 3 . . . 1 . | 1

    1 | . . 5 . . 2 . | 27 . . .11 . | 13

    2 | . 49 . . 20 | 267 . . 109 | 133

    3 | 485 . 198 | 2643 . 1079 | 1321

    . . . . .

    Rightmost 2 columns contain the values from Falzoon's answer.

  • 1 decade ago

    I don't have an answer yet, but I have a simpler equivalent problem. Use the formula S(n)=n^2(n+1)^2(2n^2+2n-1)/12. Then S(n) is a square for an integer n>0 if, and only if, 6k^2+6k+1 is a square, where k=(n-1)/3 is a positive integer. Proof: The factor of 4 in the denominator 12 doesn't affect squareness, so we need only deal with the factor of 3. By considering congruences, it is clear that if 3 divides n then S(n) is not square and similarly for 3 dividing n+1. Thus n is congruent to 1 mod 3 and can be written as 1+3k, and in this case 3 necessarily divides 2n^2+2n-1. Also, S(n) is a square if and only (2n^2+2n-1)/3 is square. Plugging in n=1+3k, we get the stated equivalence.

    Then, we need only find when this quadratic polynomial gives square values. The "Ask Dr Math" page http://mathforum.org/library/drmath/view/65773.htm... addresses this sort of question, though not giving a complete solution, but at least giving infinitely many solutions using a second order recurrence relation (and also a formula) by relating a set of solutions to solutions a Pell's equation.

  • 1 decade ago

    I found a very old post

    by Buddhist in Google Groups which gave the following

    equivalent problem: Show that

    S(n) is a square if and only if

    n = n_i, where n_i satisfies

    n_(i+2) = 10*n_(i+1) -n_i + 4.

    So I'd like to rephrase this question as: Prove this.

    You might need the formula

    S(n) = n² * (n+1)² * (2n² + 2n -1)/12

    to prove this result.

    So the first 3 solutions are

    n1 = 1, n_2 = 13, n_3 = 133.

    Update: Well, I didn't do the recursion, but

    I found a simpler way of solving this. Actually,

    it's another way of doing Duke's fine solution.

    As others have shown, we have to solve

    2n^2 + 2n -1 = 3y^2.

    or

    4n^2+4n+1 = 6y^2+3

    (2n+1)^2 -6y^2 = 3.

    Let x = 2n+1

    Then we have to solve

    x^2 -6y^2 = 3. (*)

    Let's use the arithmetic of the quadratic field Q(√6).

    Here are its properties:

    1). It is a unique factorisation domain.

    2). [1,√6] is an integral basis.

    3). The fundamental solution of

    x^2 - 6y^2 = 1 is x = 5, y = 2, so

    a fundamental unit is 5 + 2√6.

    4). Its discriminant is 24, so 3 is a square

    in this field. An element of norm 3 is 3 + √6

    and all elements of norm 3 are associated

    with 3 + √6. Further (3+√6)^2 = 3(5+ 2√6).

    All this tells us that all solutions of (*) are

    given by

    x + y√6 = (3+√6)(5+2√6)^k,

    where k is an integer.

    Letting k = 0,1,2, ...

    and n = (x-1)/2, we find the first few values

    of n to be

    n = 1, 13,133. ...

    as in Duke's table.

  • 1 decade ago

    Following on from Steiner1745, I've come up with this

    algorithm, which is not a proof, but from the recursive

    formulae, appears to show that there are an infinite

    number of solutions to S(n) = 1^5 + 2^5 + ... + n^5 = x^2.

    1. Let n_1 = 1 , n_2 = 13 , m_1 = 1 , m_2 = 11

    Also given: x_1 = 1 , x_2 = 1001

    2. Calculate: n_(i+2) = 10n_(i+1) - n_i + 4

    3. Calculate: m_(i+2) = 10m_(i+1) - m_i

    4. Calculate: x = [n_(i+2)] * [n_(i+2) + 1] * [m_(i+2)] / 2

    which leads to the following apparently infinite table -

    There are 3 columns : n_i , m_i , x_i (for i = 1, 2, 3, ...)

    1 , 1 , 1

    13 , 11 , 1001

    133 , 109 , 971299

    1321 , 1079 , 942162299

    13081 , 10681 , 913896491101

    129493 , 105731 , 886478654526101

    1281853 , 1046629 , 859883380996998799

    12689041 , 10360559 , 834085993088465707799

    125608561 , 102558961 , 809062553412431050383001

    etc.

    EDIT:

    S(n) = n^2 * (n + 1)^2 * (2n^2 + 2n - 1) / 12 = x^2

    Therefore, x = [n(n + 1) / 2] * sqrt[(2n^2 + 2n - 1) / 3]

    The recursive formula for m is for sqrt[(2n^2 + 2n - 1) / 3].

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  • 1 decade ago

    1, 13, ???

    lol, hardly what you were looking for I suspect. Sry.

    *edit*

    Hmm... I may or may not have been outdone...

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