[Guest]

N is a positive integer >=2. Determine the minimum value of N for which the root mean square of the first N positive integers (that is; 1, 2,..., N) is an integer.

Edited October 20, 2009 by K Sengupta

[Guest]

336

[superprismatic]

337

[Prof. Templeton]

337

That answer does give an integer, I can't get excel to cooperate to give me all answers leading up to that and I'm losing my patience with it. Did you guys use a different approach or program?

RMS of 1...337 = 195

[superprismatic]

That answer does give an integer, I can't get excel to cooperate to give me all answers leading up to that and I'm losing my patience with it. Did you guys use a different approach or program?

RMS of 1...337 = 195

Yes, I wrote my own,

[Guest]

Let N = 1²+2²+3²+...n

Then, the question is to find the smallest n such that root(N/n) is integer

Let X = root(N/n)

N = n(n+1)(2n+1)/6

Then, X = root((n+1)(2n+1)/6)

This means that n must be odd in order for X to be integer

Let n = 2k+1

Then, X = root((k+1)(4k+3)/3)

This means that 4k+3 must be a factor of 3; in turn, 4K must be a factor of 3

Then, let k = 3r

Overall equation now becomes,

X = root((3r+1)(4r+1))

Since 3r+1 can not be a factor of 4r+1, for X to be integer, either r must be zero or both 3r+1 and 4r+1 must be squares

i.e. 3r+1 = a² and 4r+1 = b²

(r = 0 means n = 1 which violates the condition of n>1)

Then, r = b² - a²

Substituting this above, we get

3b² = 4a² - 1

3b² = (2a+1)(2a-1)

Then a = 1 mod(3) = 3x + 1

Substituting this, we get

3b² = (6x+3)(6x+1)

or, b² = (2x+1)(6x+1)

This means that both 2x+1 and 6x+1 must be squares too

The smallest possible value of x then is 4

This means that a = 3x + 1 = 13

and b = 15

and r = 56

Then k = 56*3 = 168

n = 2k + 1 = 2*168 + 1 = 337

So, 337 is the smallest value of n for which X is integer

Being over inquisitive, I decided to find the next value of n when X could be an integer!

Well, turns out, for this, b = 209; a = 181 and r = 10920

Accordingly, k = 3r = 32760

And n = 65521

I think I know this number (65521) from somewhere... can any one help and tell me what is special about this number?

Edited October 21, 2009 by DeeGee

[Guest]

Let N = 1²+2²+3²+...n

Then, the question is to find the smallest n such that root(N/n) is integer

Let X = root(N/n)

N = n(n+1)(2n+1)/6

Then, X = root((n+1)(2n+1)/6)

This means that n must be odd in order for X to be integer

Let n = 2k+1

Then, X = root((k+1)(4k+3)/3)

This means that 4k+3 must be a factor of 3; in turn, 4K must be a factor of 3

Then, let k = 3r

Overall equation now becomes,

X = root((3r+1)(4r+1))

Since 3r+1 can not be a factor of 4r+1, for X to be integer, either r must be zero or both 3r+1 and 4r+1 must be squares

i.e. 3r+1 = a² and 4r+1 = b²

(r = 0 means n = 1 which violates the condition of n>1)

Then, r = b² - a²

Substituting this above, we get

3b² = 4a² - 1

3b² = (2a+1)(2a-1)

Then a = 1 mod(3) = 3x + 1

Substituting this, we get

3b² = (6x+3)(6x+1)

or, b² = (2x+1)(6x+1)

This means that both 2x+1 and 6x+1 must be squares too

The smallest possible value of x then is 4

This means that a = 3x + 1 = 13

and b = 15

and r = 56

Then k = 56*3 = 168

n = 2k + 1 = 2*168 + 1 = 337

So, 337 is the smallest value of n for which X is integer

Being over inquisitive, I decided to find the next value of n when X could be an integer!

Well, turns out, for this, b = 209; a = 181 and r = 10920

Accordingly, k = 3r = 32760

And n = 65521

I think I know this number (65521) from somewhere... can any one help and tell me what is special about this number?

I tried something similar but got stuck. Good work. As for the number, Google tells us that is is the largest prime number less than 2¹⁶ (16-bit number). I think its used for digit checking algorithms or something.

[superprismatic]

Let N = 1²+2²+3²+...n

Then, the question is to find the smallest n such that root(N/n) is integer

Let X = root(N/n)

N = n(n+1)(2n+1)/6

Then, X = root((n+1)(2n+1)/6)

This means that n must be odd in order for X to be integer

Let n = 2k+1

Then, X = root((k+1)(4k+3)/3)

This means that 4k+3 must be a factor of 3; in turn, 4K must be a factor of 3

Then, let k = 3r

Overall equation now becomes,

X = root((3r+1)(4r+1))

Since 3r+1 can not be a factor of 4r+1, for X to be integer, either r must be zero or both 3r+1 and 4r+1 must be squares

i.e. 3r+1 = a² and 4r+1 = b²

(r = 0 means n = 1 which violates the condition of n>1)

Then, r = b² - a²

Substituting this above, we get

3b² = 4a² - 1

3b² = (2a+1)(2a-1)

Then a = 1 mod(3) = 3x + 1

Substituting this, we get

3b² = (6x+3)(6x+1)

or, b² = (2x+1)(6x+1)

This means that both 2x+1 and 6x+1 must be squares too

The smallest possible value of x then is 4

This means that a = 3x + 1 = 13

and b = 15

and r = 56

Then k = 56*3 = 168

n = 2k + 1 = 2*168 + 1 = 337

So, 337 is the smallest value of n for which X is integer

Being over inquisitive, I decided to find the next value of n when X could be an integer!

Well, turns out, for this, b = 209; a = 181 and r = 10920

Accordingly, k = 3r = 32760

And n = 65521

I think I know this number (65521) from somewhere... can any one help and tell me what is special about this number?

You said:

"Then, X = root((k+1)(4k+3)/3)

This means that 4k+3 must be a factor of 3; in turn, 4K must be a factor of 3"

Why did you rule out that (k+1) might be a multiple of 3?

[Guest]

You said:

"Then, X = root((k+1)(4k+3)/3)

This means that 4k+3 must be a factor of 3; in turn, 4K must be a factor of 3"

Why did you rule out that (k+1) might be a multiple of 3?

If k+1 is a multiple of 3, then k must be 3t-1

Then X = root(t(12t-1))

Now, 12t-1 is never a perfect square, as a perfect square mod(12) is either 0,1,4 or 9 but not 11

This means that 12t-1 must be a factor of t in order to have X integer; which is obviously not possible either.

So, we can rule out k+1 being a factor of 3