Advanced book on Mathematics Olympiad

5.3 Diophantine Equations 275

There is a more profound way to look at this equation. Dividing through byz^2 ,we
obtain the equivalent form
(
x
z

) 2

+

(

y

z

) 2

= 1.

This means that we are supposed to find the points of rational coordinates on the unit circle.
Like any conic, the circle can be parametrized by rational functions. A parametrization
is(^1 −t
2
1 +t^2 ,

2 t

1 +t^2 ),t∈R∪ {∞}. The fractions

1 −t^2

1 +t^2 and

2 t
1 +t^2 are simultaneously rational if
and only iftitself is rational. In that caset=uvfor some coprime integersuandv. Thus
we should have

x

z

=

1 −

(u

v

) 2

1 +

(u

v

) 2 and

y

z

=

2

u

v

1 +

(u

v

) 2 ,

where again we look at the case in whichx,y, andzhave no common factor, andxand
zare both odd. Thenyis necessarily even and

y

z

=

2 uv

u^2 +v^2

.

Becauseuandvare coprime, and becauseyis even, the fraction on the right-hand side
is irreducible. Hencey= 2 uv,z=u^2 +v^2 , and consequentlyx=u^2 −v^2. Exchanging
xandy, we obtain the other parametrization. In conclusion, we have the following
theorem.

Theorem.Any solutionx, y, zto the equationx^2 +y^2 =z^2 in positive integers is of
the formx=k(u^2 −v^2 ),y= 2 kuv,z=k(u^2 +v^2 ),orx = 2 kuv,y=k(u^2 −v^2 ),
z=k(u^2 +v^2 ), wherekis an integer andu, vare coprime integers withu>vnot
both odd.

We now describe an occurrence of Pythagorean triples within the Fibonacci sequence

1 , 1 , 2 ,︸︷︷︸ 3 , 5 , 8

︸ ︷︷ ︸

, 13 , 21 , 34 , 55 , 89 , 144 , 233 ,....

Take the termsF 4 =3 andF 5 =5, multiply them, and double the product. Then take the
product ofF 3 =2 andF 6 =8. You obtain the numbers 30 and 16, and 30^2 + 162 =1156,
which is the square ofF 9 =34.
Similarly, the double product ofF 5 =5 andF 6 =8 is 80, and the product ofF 4 = 3
andF 7 =13 is 39. And 80^2 + 392 = 7921 =F 112. One more check: the double
product ofF 6 =8 andF 7 =13 is 208, the product ofF 5 =5 andF 8 =21 is 105, and
1052 + 2082 = 54289 =F 132. In general, we may state the following.