Initially 261 the perimeter of an integer-sided triangle is 75, and squares are constructed on each side. The sum of the areas of the three squares is 2009. Find the difference between the longest and shortest sides of this triangle.
Solution: Given that the sum of three numbers is 75, their average is 25.
Let $2009=(25+a)^{2}+(25+b)^{2}+(25+c)^{2}$, where $a+b+c=0$. It is easy to see that
$a^{2}+b^{2}+c^{2}=134$.
Decomposing 134 into the sum of squares of three positive integers, we find the following four cases:
(1) $a^{2}+b^{2}+c^{2}=134=121+9+4$;
(2) $a^{2}+b^{2}+c^{2}=134=100+25+9$;
(3) $a^{2}+b^{2}+c^{2}=134=81+49+4$;
(4) $a^{2}+b^{2}+c^{2}=134=49+49+36$.
For (3), let $a=9, b=-2, c=-7$. The lengths of the three sides of the triangle are
$25+9=34, 25-2=23, 25-7=18$.
Thus, the difference between the longest side and the shortest side is $34-18=16$;
Let $a=7, b=2, c=-9$. The lengths of the three sides of the triangle are
$$
25+7=32, 25+2=27, 25-9=16.
$$
Thus, the difference between the longest side and the shortest side is $32-16=16$.
However, cases (1), (2), and (4) have no solutions.
In conclusion, the difference between the longest side and the shortest side of this triangle is 16.