# 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.

### 5 Answers

- DukeLv 71 decade agoFavourite 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.

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- 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.

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- steiner1745Lv 71 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.

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- falzoonLv 71 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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- Spazzy- McGeeLv 61 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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