The numbers
,
,
, and
are not congruent numbers. In
particular,
is not. I.e., there is not rational right triangle
whose area is a perfect square, i.e., there are no ``square
triangles''.
- (60 minutes) Watch the Fermat's Last Theorem movie.
- (10 minutes) Move to computer lab / break.
- (10 minutes) Recall that we ``know''
is not a congruent
number from Monday, since
. But that
uses a deep theorem (the proof by Kolyvagin of one direction
of the Birch and Swinnerton-Dyer conjecture).
- (70 minutes) Prove Fermat's Last Theorem for
, and use this
to deduce that
is not a congruent number.
- COMPUTATION:
- Via a direct computer search, show there are no right triangles
with area
and rational side lengths
(with
the
hypotenuse) with the number of digits of the numerators and
denominators of
,
,
all
in absolute value.
- Look for solutions to
with both
rational and
. Fail to find any.
- THEORY: You definitely want to work together on these
problems, even possibly dividing up into groups and assigning parts
to different groups, and also having people search online for help
(which is allowed). Make solving all these nicely a community
effort. [Credit: This sequence of problems was originally written
by the Baur Bektemirov (the first ever winner
of the International Math Olympiad gold medal from Kazakhstan),
while he was a freshman at Harvard working with me.]
- Suppose
,
,
are relatively prime
integers with
. Then there exist integers
and
with
such that
and either
,
or
,
. Hint: Recall what we did on day 1
where we parametrized Pythagorean triples
:
- Fermat's Last Theorem for exponent
asserts
that any solution to the equation
with
satisfies
. Prove Fermat's Last
Theorem for exponent
, as follows.
- Show that if the equation
has no integer
solutions with
, then Fermat's Last Theorem for exponent
is true. (Note - the power of
is
.)
- Prove that
has no integer solutions with
as follows.
Suppose
is a solution with
minimal amongst
all solutions. Show that there exists a solution with
smaller
using the previous exercise above (consider two cases).
- Prove that 1 is not a congruent number by showing that
the cubic curve
has no rational solutions
except
and
:
- Write
and
, where
are
all positive integers and
. Prove that
, so
for some
.
- Prove that
, and substitute to see that
.
- Prove that
is a perfect square by supposing
that there is a prime
such that
is
odd and analyzing
of both sides of
.
- Write
, and substitute to
see that
. Prove that
.
- Divide through by
and deduce a contradiction to the
previous exercise.
William Stein
2006-07-07