My Math Forum Untested Equation set. (Related to Fermat Last Theorem)

 Number Theory Number Theory Math Forum

 May 2nd, 2016, 12:52 PM #1 Newbie   Joined: Apr 2016 From: Canada Posts: 16 Thanks: 0 Untested Equation set. (Related to Fermat Last Theorem) I was wondering if there's something similar to Fermat's Last Theorem but in this manner: sum series of n terms each raised to the power of n being equal to another term raised to the same power a1^n + a2^n + ... + an^n = b^n 1) a^2 + b^2 = c^2 - - - - - - - - - - - 2) a^3 + b^3 + c^3 = d^3 - - - - - - - - - - - - - - - - 3) a^4 + b^4 + c+4 + d^4 = e^4 could it be that these all have no solutions above n? so 1) no solution above 2 2) no solution above 3 3) no solution above 4 and so on? i have not tried this and not sure where to begin testing for this but just thought i'd post this. Last edited by Kensou77; May 2nd, 2016 at 12:54 PM.
 May 2nd, 2016, 08:38 PM #2 Member   Joined: Jul 2015 From: tbilisi Posts: 30 Thanks: 8 A Conjecture Of Euler. Гипотеза Эйлера — Википедия https://ru.wikipedia.org/wiki/%D0%93...B5%D1%80%D0%B0 Not true. In addition, some decisions extremely difficult. For a cube there are solutions. https://ru.wikipedia.org/wiki/%D0%97...B1%D0%B0%D1%85 I for example found such decisions. number theory - Find all integer solutions to Diophantine equation $x^3+y^3+z^3=w^3$ - Mathematics Stack Exchange For some solutions you can find there. ag.algebraic geometry - Rational solutions of the Fermat equation $X^n+Y^n+Z^n=1$ - MathOverflow
 May 3rd, 2016, 12:57 PM #3 Newbie   Joined: Apr 2016 From: Canada Posts: 16 Thanks: 0 i mean for 3 terms, there is no solution for powers above 3 4 terms no solutions for powers above 4, etc basically using the same principles of the fermat's last theorem but scaling it up. or maybe the solution cap isn't simply >n (where n is number of terms) but rather a different value on a different scale? so 1) a^2 + b^2 = c^2 - - - - - - - - - - - 2) a^3 + b^3 + c^3 = d^3 - - - - - - - - - - - - - - - - 3) a^4 + b^4 + c+4 + d^4 = e^4 but these are invalid 1) a^n + b^n = c^n n>2 - - - - - - - - - - - 2) a^n + b^n + c^n = d^n n>3 - - - - - - - - - - - - - - - - 3) a^n + b^n + c+n + d^n = e^n n>4 OR instead of linear growth, maybe 2) will cap at a higher n and 3) will also cap at a much higher n Last edited by Kensou77; May 3rd, 2016 at 12:59 PM.
 May 3rd, 2016, 08:33 PM #4 Member   Joined: Jul 2015 From: tbilisi Posts: 30 Thanks: 8 I said this hypothesis is not true. Specifically you need each equation to solve. This task is very difficult. Divination will not help here.
 May 4th, 2016, 07:04 AM #5 Senior Member   Joined: Sep 2010 Posts: 221 Thanks: 20 The hypothesis may be valid and deserving exploration if there are known examples of a^3+b^3+c^3=d^3 a^4+b^4+c^4+d^4=e^4 etc.

 Tags equation, fermat, related, set, theorem, untested

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post Rachanesamir Number Theory 2 May 13th, 2015 07:46 AM mente oscura Number Theory 10 June 6th, 2013 08:33 PM rnck Number Theory 9 August 24th, 2011 04:58 AM smslca Number Theory 4 September 14th, 2010 08:00 PM xfaisalx Number Theory 2 July 25th, 2010 04:59 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top