The following solution assumes the 3 squares are members of the set of positive integers and solely represented by such and the only function/operator to be used is the arithmetic plus (+). 

If the three squares are a, b, and c, and order doesn't matter, i.e., a+b+c is the same as a+c+b, b+a+c, b+c+a, c+a+b, and c+b+a, then there are 6 ways that 3 squares can represent 1707:

(1) 1+25+1681 = 1²+5²+41² = 1707
(2) 25+841+841 = 5²+29²+29² = 1707
(3) 49+289+1369 = 7²+17²+37² = 1707
(4) 121+361+1225 = 11²+19²+35² = 1707
(5) 121+625+961 = 11²+25²+31² = 1707
(6) 169+169+1369 = 13²+13²+37² = 1707

If each permutation is to be considered distinct, then there are 6 permuatations for each set where all 3 squares are different from each other and 3 for each set where 2 of the 3 squares are the same, for a total of 30 distinct ways.

Edited on August 23, 2012, 12:41 pm

Posted by Dej Mar on 2012-08-23 12:35:38