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Old 2011-11-14, 17:55   #1
R.D. Silverman
 
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Default Problem of the Month

Is anyone interested in starting a "problem of the month" thread?

I recently came across a good one:

For entire functions, f(z), g(z) and h(z) prove that

f^3(z) + g^3(z) = h^3(z) is impossible.
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Old 2011-11-14, 18:11   #2
davieddy
 
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Quote:
Originally Posted by R.D. Silverman View Post
Is anyone interested in starting a "problem of the month" thread?

I think my period is coming on.

Last fiddled with by davieddy on 2011-11-14 at 18:13
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Old 2011-11-14, 18:31   #3
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Pointless..
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Old 2011-11-14, 18:44   #4
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Quote:
Originally Posted by pinhodecarlos View Post
Pointless..
So mathematics is pointless? Solving math problems is pointless?

From an abstract point of view, all of 'pure' math can be considered such.
There is an old toast: "Here's to pure mathematics, may it never be of
any use to anyone".

Further, from this point of view anything that doesn't put a roof over our
head or food on the table is pointless. Football is pointless. Mastery of the
balance beam is pointless. Poetry is pointless. Fine art is pointless.


But my 'pointless' remark was in the context of doing mathematical calculations. Given that one is going to perform some calculations,
which ones should be done? It's too bad that your disdain for
reasoning ability does not let you see this.


Solving such problems shows that one has mastered mathematical
principles and reasoning and is able to apply them. Which is why,
I suppose that you think it is pointless; your prior posts seem to
indicate that you worship blind computation instead of reasoning.....
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Old 2011-11-14, 19:03   #5
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Quote:
Originally Posted by R.D. Silverman View Post
Is anyone interested in starting a "problem of the month" thread?
Sounds fun, although finding suitable problems is hard. In order not to spoil the fun for the rest of the readers of the forum, a solution should be posted by the person making the question after some time? (Here a month sounds a bit long time...).

Quote:
I recently came across a good one:

For entire functions, f(z), g(z) and h(z) prove that

f^3(z) + g^3(z) = h^3(z) is impossible.
Being clear on notation is another thing. Presumably here you mean that
"Prove that for entire functions f, g and h it is impossible to have f^3(z) + g^3(z) = h^3(z) for all z."
Also, I do not remember the standard notation in complex analysis, so could you remind me if "f^3(z)" is f\circ f \circ f(z) or f(z)^3?
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Old 2011-11-14, 19:03   #6
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Quote:
Originally Posted by pinhodecarlos View Post
.

Last fiddled with by Dubslow on 2011-11-14 at 19:04
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Old 2011-11-14, 19:18   #7
davieddy
 
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Quote:
Originally Posted by pinhodecarlos View Post
Pointless..
I sincerely hope that was just another gynocological jest.

Last fiddled with by davieddy on 2011-11-14 at 19:28
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Old 2011-11-14, 19:21   #8
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Quote:
Originally Posted by rajula View Post
Sounds fun, although finding suitable problems is hard. In order not to spoil the fun for the rest of the readers of the forum, a solution should be posted by the person making the question after some time? (Here a month sounds a bit long time...).



Being clear on notation is another thing. Presumably here you mean that
"Prove that for entire functions f, g and h it is impossible to have f^3(z) + g^3(z) = h^3(z) for all z."
Also, I do not remember the standard notation in complex analysis, so could you remind me if "f^3(z)" is f\circ f \circ f(z) or f(z)^3?
Thanks for the comment. I meant (f(z))^3. Show that the equation
is impossible to satisfy using entire functions. It is clearly meant for all z.

This is the Fermat problem for entire functions. I only have a partial solution
as yet. I understand that it was given on an exam for the 1st year grad
course in complex variables at Harvard several years ago. OTOH, it might
be very hard. When I took this course from Ahlfors he was fond of sometimes
inserting very hard or unsolved problems in problem sets........

If you think the problem is too deep for the forum, please say so.
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Old 2011-11-14, 19:25   #9
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There seems to be something missing, since f(z)=z, g(z)=2z, h(z)=cuberoot(9)z is a solution. (Or dividing by z throughout, there are lots of non-trivial constant solutions.)
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Old 2011-11-14, 19:33   #10
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Quote:
Originally Posted by Zeta-Flux View Post
There seems to be something missing, since f(z)=z, g(z)=2z, h(z)=cuberoot(9)z is a solution. (Or dividing by z throughout, there are lots of non-trivial constant solutions.)
I just passed on the problem as I heard it. You are indeed correct. I suspect that a requirement for algebraic independence may be part of the problem.

Let me check into this.... good catch. There has to me more to it.
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Old 2011-11-14, 19:36   #11
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Quote:
Originally Posted by R.D. Silverman View Post
I just passed on the problem as I heard it. You are indeed correct. I suspect that a requirement for algebraic independence may be part of the problem.

Let me check into this.... good catch. There has to me more to it.
How about a more elementary problem?

For positive integers N that possess a primitive root, prove that the
product of all of the primitive roots in Z/NZ* is 1.
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