Problem:
Suppose that the sum of the squares of two complex numbers x and y is 7 and the sum of their cubes is 10. What is the largest real value that x+y can have?
Solution:
We are given that x2+y2=7 and x3+y3=10. Because we are asked to find the sum x+y, rather than x or y individually, we are moved to rewrite these equations so as to exhibit the sum:
x2+y2x3+y3=(x+y)2−2xy=7=(x+y)3−3xy(x+y)=10(1)
This further suggests defining
s=x+y,p=xy(2)
Substituting (2) into (1) gives
s2−2ps3−3ps=7=10(3)
Solving for p in the first line of (3) and substituting the result into the second line gives
s3−3s(2s2−7)=10⟺s3−21s+20=0(4)
An obvious root of this is s=1. Dividing the cubic by s−1 gives the factorization (s−1)(s2+s−20)=0 and thence
(s−1)(s−4)(s+5)=0
Thus all values of s=x+y are real and the largest is 4.
There is a gap in this argument. Clearly any solution to (1) has led by substitution to some solution to (4), but we must also show that the largest solution for s in (4) is one which arises this way. We show more, that every solution for s in (4) arises this way. This converse is hardly automatic, as the equations are not linear.
Given any s satisfying (4), set p=2s2−7. Then (4)⇒(3). Next, for this pair (s,p), there necessarily exists a pair (x,y) satisfying (2), namely, the (possibly complex) roots of z2−sz+p=0. Finally, substituting (2) into (3) gets us back to (1). That is, there is a solution (x,y) to (1) with x+y=s.
OR
We use the method of Newton sums for determining
Sn=xn+yn,n≥0
where x and y are any two complex numbers. Setting s=x+y, p2=xy, we have again that x and y are the roots of z2−sz+p=0. We claim that
sn+2−sSn+1+psn=0,n≥0(1)
This is shown by adding xn(x2−sx+p)=xn⋅0=0 and yn(y2−sy+p)=0. Now let x and y be the desired solutions to
x2+y2=7,x3+y3=10(2)
That is, we are given S2=7 and S3=10. We seek S1=s. We also know that S0=x0+y0=2. Thus setting n=0 and then n=1 in (1), we obtain
7−s2+2p=010−7s+ps=0(3)
Thus p=2s2−7 and
10−7s+2s3−7s=0⟺s3−21s+20=0(4)
Thus, as before, s=−5,1,4 and the largest value is 4.
Also as before, the argument is not complete: every solution to (2) yields a solution to (4), but we also want the converse. The converse can be proved by methods similar to those in the first solution but a little more involved.
The problems on this page are the property of the MAA's American Mathematics Competitions