Problem:
Let a,b,c, and d be real numbers that satisfy the system of equations
a+bab+bc+caabc+bcd+cda+dababcd=−3=−4=14=30
There exist relatively prime positive integers m and n such that
a2+b2+c2+d2=nm
Find m+n.
Solution:
Answer (145):
Notice that the fourth equation implies abc⋅(−4d)=−120, and the second and third equations imply abc+(ab+bc+ca)d=abc−4d=14. Therefore abc and −4d are roots of the polynomial x2−14x−120, which are 20 and -6 . Thus {abc,−4d}={20,−6}.\
The expression ab+bc+ca is equal to ab+c(a+b)=ab−3c=−4. Also, ab⋅(−3c)=−3abc, so ab and −3c are roots of the polynomial x2+4x−3abc. Because a,b, and c are real, this polynomial must have real roots. This is not true if abc=−6. Therefore abc=20 and d=23. Then ab and −3c are roots of the polynomial x2+4x−60, which has roots -10 and 6 . Thus {ab,−3c}={−10,6}.\
Suppose ab=6. Because a+b=−3 and ab=6, it follows that a and b are roots of the polynomial x2+3x+6. But this polynomial has no real roots. Therefore ab=−10 and c=−2. Then a and b are roots of the polynomial x2+3x−10, which has roots -5 and 2 . Thus
a2+b2+c2+d2=(−5)2+22+(−2)2+(23)2=4141
The requested sum is 141+4=145.
The problems and solutions on this page are the property of the MAA's American Mathematics Competitions