26 2 Algebraand the last written by the second author of the book, and the second given at a Soviet
Union college entrance exam and suggested to us by A. Soifer.
Example.Solve in real numbers the system of equations
( 3 x+y)(x+ 3 y)√
xy= 14 ,
(x+y)(x^2 + 14 xy+y^2 )= 36.
Solution.By substituting√
x=u,√
y=v, we obtain the equivalent form
uv( 3 u^4 + 10 u^2 v^2 + 3 v^4 )= 14 ,
u^6 + 15 u^4 v^2 + 15 u^2 v^4 +v^6 = 36.
Here we should recognize elements of the binomial expansion with exponent equal to 6.
Based on this observation we find that
36 + 2 · 14 =u^6 + 6 u^5 v+ 15 u^4 v^2 + 20 u^3 v^3 + 15 u^2 v^4 + 6 uv^5 +v^6and36 − 2 · 14 =u^6 − 6 u^5 v+ 15 u^4 v^2 − 20 u^3 v^3 + 15 u^2 v^4 − 6 uv^5 +v^6.Therefore,(u+v)^6 =64 and(u−v)^6 =8, which impliesu+v=2 andu−v=±√
2
(recall thatuandvhave to be positive). Sou= 1 +√ 2
2 andv=^1 −√ 2
2 oru=^1 −√ 2
2
andv= 1 +√
2
2. The solutions to the system are(x, y)=(
3
2
+
√
2 ,
3
2
−
√
2
)
and (x, y)=(
3
2
−
√
2 ,
3
2
+
√
2
)
.
Example.Given two segments of lengthsaandb, construct with a straightedge and a
compass a segment of length^4√
a^4 +b^4.
Solution.The solution is based on the following version of the Sophie Germain identity:
a^4 +b^4 =(a^2 +√
2 ab+b^2 )(a^2 −√
2 ab+b^2 ).
Write√ (^4) a (^4) +b (^4) =
√√
a^2 +√
2 ab+b^2 ·√
a^2 −√
2 ab+b^2.Applying the law of cosines, we can construct segments of lengths
√
a^2 ±√
2 ab+b^2
using triangles of sidesaandbwith the angle between them 135◦, respectively, 45◦.
On the other hand, given two segments of lengthsx, respectively,y, we can construct
a segment of length√
xy(their geometric mean) as the altitudeADin a right triangle
ABC(∠A= 90 ◦) withBD=xandCD=y. These two steps combined give the
method for constructing^4√
a^4 +b^4.