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        riddles >> easy >> Sum of 999 perfect squares
        
(Message started by: NickH on Jan 27^th, 2004, 3:47pm)

Title: Sum of 999 perfect squares
Post by NickH on Jan 27^th, 2004, 3:47pm

Can the sum of 999 consecutive squares of integers ever be a perfect power of an integer?

Example: are there integers m, r > 1, such that 1005² + ... + 2003² = m^r?

Title: Re: Sum of 999 perfect squares
Post by Eigenray on Jan 27^th, 2004, 11:38pm

::[hide]Nope[/hide]::
Let f(n) = [sum]_k=nⁿ⁺⁹⁹⁸ k².
[hide]f(n+1)-f(n) = (n+999)²-n² = 999(2n+999) == 0 mod 3³.
Therefore f(n) == f(1) = 999(1000)(1999)/6 == 18 mod 27.
Now, if m^r = f(n) = 27k + 18 = 3²(3k+2), then r <= 2.
But if r = 2, then we have (m/3)² = 3k+2, which is impossible[/hide].

Title: Re: Sum of 999 perfect squares
Post by NickH on Jan 28^th, 2004, 12:48pm

Nice solution!

I had something similar:

If m^r = (n - 499)² + ... + (n + 499)² = 999n² + 2(1² + ... + 499²) = 999n² + 499*500*999/3 = 333(3n² + 499*500), then 3|m^r, but not 3²|m^r. Contradiction.

Title: Re: Sum of 999 perfect squares
Post by Eigenray on Jan 28^th, 2004, 4:49pm

on 01/28/04 at 12:48:46, NickH wrote:

If m^r = [...] = 333(3n² + 499*500), then 3|m^r, but not 3²|m^r.

But 3² does divide m^r.

Title: Re: Sum of 999 perfect squares
Post by NickH on Jan 29^th, 2004, 1:15am

Oops! I see I carelessly generalised from the problem I'd seen.

The proof I gave is valid for 99 or 9999 or 10²ⁿ - 1 consecutive squares, but not for 10^2n-1 - 1 consecutive squares.

Thanks Eigenray.