[Guest]

Find the first integer n > 1 such that the average of

1^2, 2^2, 3^2, . . . , n^2

is itself a perfect square.

[Guest]

337 gives 195^2

Average of the first n squares is (1/6)(n+1)(2n+1)

The first n that gives a perfect square besides n=1 is n=337

Edited December 6, 2009 by mmiguel1

[Guest]

for integer 24, I get a sum of squares upt o and including 24 = 4900, which is a square of 70. I guess I could do a quick check to verify that...I do have a tendency to make typos...

[Guest]

oops I did: sums of squares are squares! we're looking for the average value of squares is a square: yeah,integer 337 yields a square avg = 38025.0, which is a square of 195

[Guest]

oops I did: sums of squares are squares! we're looking for the average value of squares is a square: yeah,integer 337 yields a square avg = 38025.0, which is a square of 195

I did the same thing on my first try, haha.

Edited December 6, 2009 by mmiguel1

[Guest]

I did the same thing on my first try, haha.

that's such an awesome coincidence! I like these math puzzles because they give me a reason to write code in Ruby. Sounds like you actually do some mathematical figuring....

[Guest]

that's such an awesome coincidence! I like these math puzzles because they give me a reason to write code in Ruby. Sounds like you actually do some mathematical figuring....

Not really, haha.

I had known the formula for the sum of squares already and that it had n as a factor.

I wrote a little python script to find the first n that gives a perfect square using the modified formula. Ruby is great too though!

Edited December 6, 2009 by mmiguel1

[Guest]

I used excel and scrolled down to find the answer and got the same.

[Guest]

337 determined from Excel calculations

[random7]

Find the first integer n > 1 such that the average of

1^2, 2^2, 3^2, . . . , n^2

is itself a perfect square.

I have reduced the problem to trying to find a natural-number "n" such that:

(n^2)/3 + n/2 + 1/6 = m^2 (for a natural-number "m")

but I am stuck in trying to find an expression for the square-root of the LHS (so I can then find the least "m" which satisfies the equation).

Does anyone know how to do this?

[Guest]

I have reduced the problem to trying to find a natural-number "n" such that:

(n^2)/3 + n/2 + 1/6 = m^2 (for a natural-number "m")

but I am stuck in trying to find an expression for the square-root of the LHS (so I can then find the least "m" which satisfies the equation).

Does anyone know how to do this?

Computer! Lol

Write a program or use a spreadsheet to determine the square root for various n.

The first one that comes up as a whole number corresponds to your answer.

In python this will do the trick:

(replace ~ with spaces or tabs)

import math;

n=2;

while 1:

~m = math.sqrt((1/6.0)*(n+1)*(2*n+1));

~if m == int(m):

~~print n, str(m)+"^2";

~~break;

~n += 1;

Edited December 9, 2009 by mmiguel1

[random7]

Computer! Lol

Write a program or use a spreadsheet to determine the square root for various n.

The first one that comes up as a whole number corresponds to your answer.

In python this will do the trick:

(replace ~ with spaces or tabs)

import math;

n=2;

while 1:

~m = math.sqrt((1/6.0)*(n+1)*(2*n+1));

~if m == int(m):

~~print n, str(m)+"^2";

~~break;

~n += 1;

Yeah, computer programs are good for solving problems... But puzzles are meant to be solved using logic, deduction, and all that jazz... I was wanting to find the answer using logic alone, not brute force = ) so my question still stands. Thanks for your response though, it would of been helpful otherwise.

Does OP know how to solve this puzzle?