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The Fermat's Last Theorem was proved by Andrew Wiles in 1995. It states that no three positive integers $x$, $y$, and $z$ can satisfy the equation $x^n+y^n= z^n$ for any integer value of $n$ greater than two.

I wonder if there is general solution for $a^n+b^n+c^n = d^n$ , $n>4$ or not?

$a,b,c,d,n$ are positive integers

$$a^n+b^n+c^n = d^n$$

For $n=2$, there is a solution $$1^2+2^2+2^2=3^2$$

For $n=3$, there is a solution $$3^3+4^3+5^3=6^3$$

I saw somewhere that For $n=4$, there is a solution $$95800^4 + 217519^4 + 414560^4 = 422481^4$$

Is there proof for the general problem that there is no solution for $n>4$?

Do you know counterexamples to disprove the generalization for $n>4$?

Thanks for answers

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1 Answer 1

up vote 8 down vote accepted

From the second edition of Unsolved Problems in Number Theory by Richard K. Guy, section D1, Euler conjectured that your equation should be impossible for $n \geq 4.$ For $n=4,$ the first counterexample was by Noam Elkies, a parametric solution was given by Dem'janenko, and your solution above was found by Roger Frye. No explicit reference for Frye...There is a third edition, there it starts on page 209. I don't think anyone has any solutions for $n>4.$

To be specific, Euler thought that you needed four fourth powers to add up to another fourth power, five fifth powers to add up to another fifth power, and so on.

OK, we have $$ 27^5 + 84^5 + 110^5 + 133^5 = 144^5, $$ by Lander and Parkin (1966). I don't think three fifth powers is going to work. Might try six or five or four sixth powers, or seven or six or five seventh powers if you want to stick with prime exponents. Note that, by Fermat's little theorem, for sixth powers , all but one of the summands will need to be divisible by 7. That is, if some $a \neq 0 \pmod 7$ then $a^6 \equiv 1 \pmod 7.$ You have fewer than seven summands, so the left hand side will not be divisible by 7, so the right hand side will be exactly $1 \pmod 7.$ Something of a timesaver. To compare, look at the solution in the original question for $n=4,$ two out of three numbers on the left hand side are divisible by 5.

Alright, as of the second edition (1994) there were no known solutions for $n$ $n$th powers adding up to another $n$th power for $n \geq 6.$ So, even what Euler had intended as the problem had no known solutions. Of course, computers are much faster now...

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Thanks a lot for the references. I have been using them while searching the subject. I wonder why the method that Wiles used to prove for Fermat's Last Theorem does not work for the problem I asked. If you know the proof of Fermat's Last Theorem, could you please advice about difficulties and differences of the problem I asked from FLT? Thanks –  Mathlover Aug 12 '13 at 12:38
    
@Mathlover, my advice would be for you to find good presentations of the proof of FLT for exponent 5, begin with list at en.wikipedia.org/wiki/… and see how things go when trying to prove that the sum of three fifth powers cannot be another fifth power. At the same time, you could run a computer search to find such an equation. –  Will Jagy Aug 12 '13 at 20:36
    
@Mathlover, $n=6$ is likely to be easier to deal with, en.wikipedia.org/wiki/… –  Will Jagy Aug 12 '13 at 21:51
    
@WillJagy, Actually, for $n$ $n$th powers equal to an $n$th power, while there is none for $n=6$, there is for $n=8$. See Eight Powers. If we allow negative integers, then there are for $n=7,9$ as well. –  Tito Piezas III Aug 18 '13 at 1:44

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