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 Number Theory Number Theory Math Forum

 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 $\displaystyle 40^3+17^3+2^3=41^3$ Thanks from Maschke and topsquark 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
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 Originally Posted by Loren 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.
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/ Thanks from Loren 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". Thanks from Loren Tags >3, >3, fermat, n>0, n>0, terms Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Loren Number Theory 0 March 8th, 2017 09:46 PM complicatemodulus Number Theory 47 April 17th, 2015 05:34 AM McPogor Number Theory 0 July 28th, 2014 10:58 AM fakesmile Calculus 0 June 2nd, 2009 05:36 AM circum Number Theory 1 June 15th, 2008 05:20 PM

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