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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. 

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