-2

When you have an equation such as 3271^11 + 5012^11 = 5193^11, what is the best way to prove it is false?

FΣynman
  • 1
  • If you're ok with using bigger results, if it were true it would be an example of Fermat's Last Theorem holding, which is known to have no non-trivial solutions for $n\geq 3$ (your case is $n = 11$).

    Besides that, subtracting $5012^{11}$ from both sides might be useful, as: $$x^n-y^n = (x-y)\left(\sum_{i = 0}^{n-1}x^iy^{n-1-i}\right)$$ This will let you look at possibly an easier problem.

     – Mark Schultz-Wu Nov 07 '16 at 01:01
  • You can manipulate with the inequality symbol != in precisely the same way you can manipulate the equality symbol, adding or multiplying by the same number preserves truth value. Thus you reduce it to something like 1 is not equal to 0, a tautology. – Jacob Wakem Nov 07 '16 at 01:06

2 Answers2

2

Well, will you accept this?

$(10x+1)^{11}$ will have last digit $1$.

$(10y+2)^{11}$ will have last digit $8$ (because $2^{11}=2048$)

And $(10z+3)^{11}$ will have last digit ...um, lets, see, 3,9,27,21,3,9,7,1,3,9,7... $7$. It will have a last digit of $7$.

And $1+8 \ne 7$.

fleablood
  • 124,253
  • good one, but how to avoid "counting". i.e. what if there would stand $1234^{11}+12345^{11}=123456^{11}$ I mean, how to avoid counting in $3^{11}$. Simply 11 modulo 4? – SAJW Nov 07 '16 at 01:16
  • 1
    Last digit of 4^11 is 4 and 6^11. So that's not a counter example but 1^11 + 5^11 will have last digit 6 which would be consistant with 6^11. So you have to do something else. But note last digits always go in patterns of four so you can only have a 4x10x10 f possible last digit combos, most of which fail. – fleablood Nov 07 '16 at 02:09
1

First note that if $a$ and $b$ are integers, $a^n-b^n$ is always divisible by $a-b$. Thus, $5193^{11}-5012^{11}$ is divisible by $5193-5012=181$. It’s not hard to check that $181$ is prime, so $3271^{11}$ is divisible by $181$ if and only if $5193$ is divisible by $181$. Is it?

Brian M. Scott
  • 616,228