Find all natural numbers that can be expressed in a unique way as a sum of five or less perfect squares.
Problem
Source: CroatianTST 2006 and 18-th KoreanMO 2005(Final Round)
Tags: number theory proposed, number theory
09.05.2007 21:10
Only 0, because 1=02+12=12+02.
10.05.2007 07:58
But I see that 2 also is a solution of problem. Note: 1)Parts of sum must are positive integers. 2)12+22 and 22+12 are same. Now try again!
10.05.2007 10:32
Two well known and already posted theorems: a) every integer can be written as sum of four squares. b) every integer n>N (with e.g. N=169) can be written as sum of five nonzero squares. Thus every big integer can be written in two ways, so we are left to check some small numbers, done.
10.05.2007 13:11
ZetaX wrote: b) every integer n>N (with e.g. N=169) can be written as sum of five nonzero squares. Wow, please post that link here, thanks
10.05.2007 13:39
http://www.mathlinks.ro/Forum/viewtopic.php?t=121595
12.05.2007 11:04
Hello, ZetaX wrote: Two well known and already posted theorems: a) every integer can be written as sum of four squares. b) every integer n>N (with e.g. N=169) can be written as sum of five nonzero squares. Thus every big integer can be written in two ways, so we are left to check some small numbers, done. I think that if it is possible to use four squares theorem, it's useless to use b) and check up to 169. There is something simpler : For n≥16, if n can be written as a2+b2+c2+d2 with either 1, either 2, either 3, either 4 not in the set {a,b,c,d}, then n can be written as u2+v2+w2+t2+x2 where x is the missing number (1,2,3 or 4) and (u2,v2,w2,t2) the 4 or less squares of expression of n−x2. Because x is not in in the set (a,b,c,d), the two expressions are different. So we have just to check numbers 1 to 15 plus the number 30=12+22+32+42. 1=1 is OK 2=1+1 is OK 3=1+1+1 is OK 4=1+1+1+1=4 is NOK 5=4+1=1+1+1+1+1 is NOK 6=4+1+1 is OK 7=4+1+1+1 is OK 8=4+4=4+1+1+1+1 is NOK 9=9=4+4+1 is NOK 10=9+1=4+4+1+1 is NOK 11=9+1+1=4+4+1+1+1 is NOK 12=9+1+1+1=4+4+4 is NOK 13=9+1+1+1+1=4+4+4+1 is NOK 14=9+4+1=4+4+4+1+1 is NOK 15=9+4+1+1 is OK 30=16+9+4+1=25+4+1 is NOK And the requested set is {1,2,3,6,7,15} (+ 0 is 0 is to be considered) A question now is : can this problem be solved without using four squares theorem ? -- Patrick