March 19th, 2018, 09:11 PM  #1 
Senior Member Joined: May 2015 From: Arlington, VA Posts: 397 Thanks: 27 Math Focus: Number theory  Fermat's Last for >3 terms, n>0
Fermat's Last Theorem states that no three positive integers a, b, and c satisfy the equation a^n + b^n = c^n for any integer value of n greater than 2. Is there a solution for four or more such terms with an integer n>0? 
March 20th, 2018, 03:20 AM  #2 
Senior Member Joined: Feb 2010 Posts: 704 Thanks: 138 
$\displaystyle 40^3+17^3+2^3=41^3$

March 20th, 2018, 11:32 AM  #3 
Senior Member Joined: May 2015 From: Arlington, VA Posts: 397 Thanks: 27 Math Focus: Number theory 
I meant to write "is there always a solution to four or more such terms for each integer n>0?"

March 20th, 2018, 01:04 PM  #4 
Global Moderator Joined: May 2007 Posts: 6,704 Thanks: 670 
$3^3+4^3+5^3=6^3$

March 20th, 2018, 05:07 PM  #5 
Senior Member Joined: May 2015 From: Arlington, VA Posts: 397 Thanks: 27 Math Focus: Number theory 
In general, does there always exist at least one solution for given n where a^n+b^n+c^n+ ... +d^n=z^n with [at least] n+1 terms, having a, b, c, d, z and n integers greater than zero? __________ This is an extension of Fermat's Last Theorem. 
March 21st, 2018, 12:30 PM  #6 
Global Moderator Joined: May 2007 Posts: 6,704 Thanks: 670  Why "at least"? Why not less?

March 21st, 2018, 03:59 PM  #7 
Senior Member Joined: May 2015 From: Arlington, VA Posts: 397 Thanks: 27 Math Focus: Number theory 
Including "at least" would have put fewer constraints on my equation. My guess is, that would make a less defined equation with "looser" results. The bracket showed my initial uncertainty. But here, without that bracketed phrase, I am looking for possible unique results for any given n>3. It has been shown previously that there are multiple cases where n=3. My equation has a minimum of n=1 terms separated by an equal sign. "Less than" n+1 terms, where n=0, would not indicate an equation. Fermat's Last Theorem eliminated all possible n=2 for its 3 terms, with the exception of n=1 and n=2. I included the symmetry I did for my hypothesis, so n is fundamentally related to the number of terms. mathman, your suggestion is worthwhile; I was trying to narrow the field of answers by considering a greater symmetry for n>3, with perhaps singular answers for each n exponents with their n+1 terms. __________ Does there always exist at least one solution for given n where a^n+b^n+c^n+ ... +d^n=z^n with n+1 terms, having a, b, c, d, z and n integers greater than zero? 
March 24th, 2018, 12:53 PM  #8 
Global Moderator Joined: May 2007 Posts: 6,704 Thanks: 670 
I posted this question on another forum, where you should get more responses. https://math.stackexchange.com/ 
March 25th, 2018, 09:48 AM  #9 
Senior Member Joined: May 2015 From: Arlington, VA Posts: 397 Thanks: 27 Math Focus: Number theory 
Much appreciation for the site, mathman. How do I find my post there?

March 25th, 2018, 12:26 PM  #10 
Global Moderator Joined: May 2007 Posts: 6,704 Thanks: 670 
At the end of the list you can click on popular tags. Next page click on users. The search for my name (herb steinberg). You will see the title "Generalized Fermat's Last Theorem".


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>3, >3, fermat, n>0, n>0, terms 
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