Problem:
Find the number of triples of nonnegative integers (a,b,c) satisfying a+b+c=300 and
a2b+a2c+b2a+b2c+c2a+c2b=6,000,000.
Solution:
Adding 3abc to the second equation gives
6,000,000+3abc=a2b+a2c+b2a+b2c+c2a+c2b+3abc=(a+b+c)(ab+bc+ca)=300(ab+bc+ca).
Hence 2,000,000+abc=100(ab+bc+ca). On the other hand,
(a−100)(b−100)(c−100)=abc−100(ab+bc+ca)+1002(a+b+c)−1003=abc−100(ab+bc+ca)+1002⋅300−1003=abc−100(ab+bc+ca)+2,000,000.
Combining the two obtained relations gives
(a−100)(b−100)(c−100)=0.
In particular, one of a,b, and c must be 100. Conversely, if one of the variables, say c, is equal to 100 and a+b=300−c=200, then
a2b+a2c+b2a+b2c+c2a+c2b=ab(a+b)+100(a2+b2)+c2(a+b)=100(a+b)2+200c2=6,000,000.
Therefore the triples that satisfy the given equation are precisely the triples for which one of the variables is equal to 100. There are 201 valid triples with c=100 and a+b=200. Likewise, there are 201 valid triples with each of a=100 and b=100. However, the triple (100,100,100) has been counted three times, so the requested number of triples is 3⋅201−2=601.
OR
Grouping terms of the second equation gives
6,000,000=a2b+a2c+b2a+b2c+c2a+c2b=(a2b+b2a)+(a2c+c2a)+(b2c+c2b)=ab(a+b)+ac(a+c)+bc(b+c)=ab(300−c)+ac(300−b)+bc(300−a)=300(ab+bc+ca)−3abc.
Hence abc=100(ab+bc+ca)−2,000,000.
Let D=ab+bc+ca. Then a,b, and c are solutions of the cubic equation
x3−300x2+Dx−(100D−2,000,000)=0.
By inspection, x=100 is a solution of this equation, and thus one of a,b, or c is equal to 100. The rest of the solution then proceeds as in the first solution.
The problems on this page are the property of the MAA's American Mathematics Competitions