Dado um triângulo ABC, marca-se um ponto D no lado AB. Mostre que CD é menor do que o comprimento de um dos lados AC ou BC.
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We have:
Thus:
Adding
gives:
Factoring gives:
Thus, we have:
or
.
Note that because
, there will be no solutions 
of the form 
for positive integer 
and solution 
. So we find the basic values of 
giving 
.
For with positive , 
(this can be seen easily from the unit circle. But 
, a contradiction. So we examine the case in which .
In this case,
. So this is the only solution to this equation.
Substituting this back in, we have
, as desired. So this is a valid solution.
Thus,.
Thus:
Adding
Factoring gives:
Thus, we have:
or
Note that because
For
In this case,
Substituting this back in, we have
Thus,