Patrick Snow of Dame Alice Owens School, London,
Marcos Charalambides from Cyprus and Andrei
Lazanu from Romania sent solutions to this
problem.
The famous golden ratio is
g=
Ö5 + 12
. Patrick and Andrei showed that
g2 =
(Ö5 + 1 )24
=
(6 + 2Ö5)4
=
(Ö5 + 3)2
= g + 1
.
Marcos considered the equation x2 = x + 1. By
completing the square,
(x - 1/2)2 = 5/4
he
obtained the solutions
x=
1 ±Ö52
.
So g is one solution of this equation and
hence g2=g+1.
Both Andrei and Marcos then experimented a bit to
get familiar with the problem...
Multiplying by g
g2
= g+1
g3
= g2 + g = 2g + 1
g4
= 2g2 + g = 3g + 2
g5
= 3g2 + 2g = 5g +3
g6
= 5g2 + 3g = 8g + 5
¼
.
Marcos then made the conjecture that an+1 = an + an-1 and these coefficients are the
Fibonacci sequence. Similarly for the bn and
went on to prove this conjecture using the method
of mathematical induction.
Patrick used the result g2=g+1 to find a
pattern in the coefficients for gn = ang +bn as follows. Multiplying by g gives
Thus an+1=an+bn and
bn+1=an. Hence an+1=an + an-1.
This is the Fibonacci sequence. Since the first
terms are a1=1 and a2=1 so
gn=ang+an-1 where ak is the k-th term
of the Fibonacci sequence.
Then Patrick went on to prove by induction that
an=
an-bnÖ5
where
a and b are the solutions of the
quadratic equation x2 = x + 1.
Now for n=1
a- b =
1+Ö52
-
1-Ö52
=Ö5
and dividing this by Ö5 gives the value of a1=1 showing that the result is true for n=1.
For n=2
a2 - b2Ö5
=
(a+ 1)- (b+1)Ö5
= 1
as above
giving the value of a2=1 showing that the
result is true for n=2.
Assume the result for n=k and n=k-1 and using
the fact that a2 = a+1 and
b2=b+1 then
ak+1
= ak + ak-1
=
ak-1(a+1)-bk-1(b+1)Ö5
=
ak+1-bk+1Ö5
Hence by the axiom of induction the result is
proved. Thus