20111114, 17:55  #1 
"Bob Silverman"
Nov 2003
North of Boston
2^{3}·937 Posts 
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. 
20111114, 18:11  #2  
"Lucan"
Dec 2006
England
2×3×13×83 Posts 
OMG
Quote:
I think my period is coming on. Last fiddled with by davieddy on 20111114 at 18:13 

20111114, 18:31  #3 
"Carlos Pinho"
Oct 2011
Milton Keynes, UK
1001111110111_{2} Posts 
Pointless..

20111114, 18:44  #4 
"Bob Silverman"
Nov 2003
North of Boston
2^{3}·937 Posts 
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..... 
20111114, 19:03  #5  
"Tapio Rajala"
Feb 2010
Finland
3^{2}·5·7 Posts 
Quote:
Quote:
"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 or ? 

20111114, 19:18  #7 
"Lucan"
Dec 2006
England
2×3×13×83 Posts 
I sincerely hope that was just another gynocological jest.
Last fiddled with by davieddy on 20111114 at 19:28 
20111114, 19:21  #8  
"Bob Silverman"
Nov 2003
North of Boston
2^{3}·937 Posts 
Quote:
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. 

20111114, 19:25  #9 
May 2003
7·13·17 Posts 
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 nontrivial constant solutions.)

20111114, 19:33  #10  
"Bob Silverman"
Nov 2003
North of Boston
1D48_{16} Posts 
Quote:
Let me check into this.... good catch. There has to me more to it. 

20111114, 19:36  #11  
"Bob Silverman"
Nov 2003
North of Boston
16510_{8} Posts 
Quote:
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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