Problem:
Let a,b, and c be positive integers with a≥b≥c such that
a2−b2−c2+aba2+3b2+3c2−3ab−2ac−2bc=2011=−1997
What is a?
Answer Choices:
A. 249
B. 250
C. 251
D. 252
E. 253
Solution:
Adding the two equations gives
2a2+2b2+2c2−2ab−2bc−2ac=14
so
(a−b)2+(b−c)2+(c−a)2=14.
Note that there is a unique way to express 14 as the sum of perfect squares (up to permutations), namely, 14=32+22+12. Because a−b,b−c, and c−a are integers with their sum equal to 0 and a≥b≥c, it follows that a−c=3 and either a−b=2 and b−c=1, or a−b=1 and b−c=2. Therefore either (a,b,c)=(c+3,c+1,c) or (a,b,c)=(c+3,c+2,c). Substituting the relations in the first case into the first given equation yields 2011=a2−c2+ab−b2=(a−c)(a+c)+(a−b)b=3(2c+3)+2(c+1). Solving gives (a,b,c)=(253,251,250). The second case does not yield an integer solution. Therefore a=(E)253.
The problems on this page are the property of the MAA's American Mathematics Competitions