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Challenge

Name: Anonymous 2009-01-27 19:39

Problem: Find a positive integer n such that n^2 is the sum of twenty five non-zero squares, and give the squares.

First correct answer wins a cookie! :D

Name: Anonymous 2009-01-27 19:49

/sci/ - Help Retards With Their Math Homework So They Learn Neither The Material Nor Critical Thinking Skills For Themselves And They End Up Failing Higher Level Courses

Name: Anonymous 2009-01-27 23:34

the answer is 5.
1 + 1 + 1 + 1 etc.

Name: Anonymous 2009-01-28 1:46

25((>>2-1)^2)

Name: Anonymous 2009-01-28 8:13

surely this works for any n s.t 5|n.

Just take n = 5m,

then add together 25 copies of m^2.

If you want the squares to be distinct it'd be a little different.

Name: Anonymous 2009-01-28 11:45

>If you want the squares to be distinct it'd be a little different.

OP here:  Lol yeah I meant to say "different squares" actually.

Oh well, u can all have cookies anyway. ;)

http://www.zuzzys.com/images/cookies.jpg

Name: Anonymous 2009-01-28 19:38

I'm not certain how you would approach this beyond trial and error (Obviously could use some sort of modular arithmetic to speed it up, but can't see how it would provide a nice solution)

Is there actually a "nice" answer, or is it some bullshit you have to spot?

Name: Anonymous 2009-01-28 21:05

I found one, but it has three and a half million digits. (seriously)  Recursive pythagorean triples.

3^2 + 4^2 = 5^2
5^2 + [(5^2-1)/2]^2 = [(5^2+1)/2]^2 = 13^2
13^2 + [(13^2-1)/2]^2 = [(13^2+1)/2]^2 = 85^2
85^2 + [(85^2-1)/2]^2 = [(85^2+1)/2]^2 = 3613^2

etc.

So 3613^2 = 3^2 + 4^2 + 12^2 + 84^2 + 3612^2, and with the same method you can construct squares that are sums of however many squares you want.  I ran it in mathematica, and the number of digits doubles with each step.

Name: Anonymous 2009-01-29 18:16

Now do it with cubes.

Name: mo‮8pE! to‬pui‮ !1uHaijp7IU!n9e4aOufPFUPnvI 2009-01-31 5:21

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