Re: 3x^2y^2+12xy^2+16y^2=z^3
- From: quasi <quasi@xxxxxxxx>
- Date: Thu, 05 Jul 2007 00:08:00 -0500
On Wed, 04 Jul 2007 16:26:53 EDT, tommy1729 <tommy1729@xxxxxxxxx>
wrote:
Firstly, note that your reply lacks the correct attributions as to who
said what. I've asked you several times to try to learn how to quote
prior context correctly, but there's been no improvement.
For this reply, I'll try to fix the context by hand, but it _can_ be
done automatically if you learn how to use the options provided by
your posting program.
Please make an effort.
Ok, so the math ...
Vincenzo Librandi wrote:
Why isn't
3x^2y^2+12xy^2+16y^2=z^3 ?
Tommy replied:
the left side must be devisable bye y^2.
therefore z^3 must also be devisable bye y^2.
therefore z must be devisable bye y.
Quasi objected:
No, not correct.
For example, for any prime p, if a=p^4 and b=p^3,
then a^2 divides b^3 but a does not divide b.
Tommy responds:
ah yes of course a does not devide b if b is larger than a.
Quasi now corrects ...
??
Perhaps you meant to say
a does not divide b if a is larger than b
which is at least correct if a,b are positive integers.
Tommy continues ...
otherwise it seems correct , not ?
Quasi responds ...
No.
The statement
"If a^2 divides b^3 then a divides b"
can be falsified with a>b and can also be falsified with a<b.
Counterexample #1 (with a>b):
If a=8, b=4 then a^2 divides b^3 but a does not divide b.
Counterexample #2 (with a<b):
If a=8, b=12 then a^2 divides b^3 but a does not divide b.
quasi
.
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